Important Questions

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 8 Rectification of Errors Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 8 Rectification of Errors

1. Answer in One Sentence:

Question 1.
What is an Arithmetical error?
Answer:
An error committed to totaling the amount columns of journal and ledger is called arithmetical error.

Question 2.
What do you mean by one-sided errors?
Answer:
One-sided errors are those errors that either affect the debit or credit aspect of the transaction. It affects the agreement of the trial balance.

Question 3.
What do you mean by two-sided errors?
Answer:
Two-sided errors are those errors that affect both the aspects of a transaction i.e. debit and credit. It does not affect the agreement of the trial balance.

Question 4.
What is a suspense account?
Answer:
An account that is opened to transfer the difference in the totals of the trial balance is known as a suspense account.

Question 5.
What are accounting errors?
Answer:
Mistakes or errors committed while writing the books of accounts are known as accounting errors.

2. Give one word/term or phrase for each of the following statements.

Question 1.
Transactions remained to be recorded at all in the books of account.
Answer:
Errors of omission

Question 2.
Errors are always rectified by passing rectification entries.
Answer:
Two-sided errors

Question 3.
Errors that affect the debit and credit side unequally.
Answer:
One-sided errors

Question 4.
Errors that affect the debit and credit side equally.
Answer:
Two-sided errors

Question 5.
An error in which the transaction is entered in the original book but not posted into the ledger.
Answer:
The error of posting/partial omission

Question 6.
Error in the process of transferring the entry from original books into the ledger.
Answer:
Error of posting

Question 7.
Errors can be rectified without passing rectification entries.
Answer:
One-sided errors

3. Select the most appropriate alternative from those given below and rewrite the sentence.

Question 1.
Wages paid for the installation of machinery wrongly debited to wages account is an error of ___________
(a) omission
(b) principle
(c) commission
(d) duplication
Answer:
(b) principle

Question 2.
If the trial balance shows a short credit the suspense account will have a ___________ balance.
(a) debit
(b) zero
(c) credit
(d) nil
Answer:
(c) credit

Question 3.
If the trial balance does not agree the difference of the trial balance is placed in ___________ account.
(a) Personal
(b) Suspense
(c) Rectification
(d) Real
Answer:
(b) Suspense

Question 4.
Errors which compensate the effect of each other are called ___________ errors.
(a) compensating
(b) one-sided
(c) two sided
(d) clerical
Answer:
(a) compensating

Question 5.
One sided errors are disclosed by ___________
(a) Trial Balance
(b) Suspense Account
(c) Journal
(d) Ledger Account
Answer:
(a) Trial Balance

4. State whether the following statements are True or False with reasons.

Question 1.
Errors of principle are not disclosed by the Trial Balance.
Answer:
This statement is True.
The error of Principle is those where some basic principles of bookkeeping and accountancy are not Properly Followed while recording a business transaction. Such errors can be rectified by passing journal entries.
Eg.: Capital expender showed as revenue expenditure or vice versa. It won’t affect the Trial balance. If will agreed.

Question 2.
Transaction not recorded in the books is an error of principle.
Answer:
This statement is False.
Transaction not recorded in the books is an error of omission and not an error of principle.

Question 3.
If the purchase hook is undermasted, the purchase account is debited in rectification.
Answer:
This statement is True.
Purchases are expenses. All expenses are debited as per the rule of nominal account. When purchases are underacted they can be corrected by debiting in rectification.

Question 4.
When a transaction is not recorded according to the principles of book-keeping the error is said to be an error of principle.
Answer:
This statement is True.
The error of Principle is those where some basic principles of bookkeeping and accountancy are not properly followed while recording a business transaction.
Eg.: Capital expenditure is shown as revenue expenditure or Vice-a-Varsa and they are called an error of principle.

Question 5.
The error of omission is disclosed by the Trial Balance.
Answer:
This statement is False.
The complete omission of a transaction will bot disclosed by a trial balance. Trial balance, balances will be agreed with such errors. So the omission of transaction or an error of omission will not be disclosed by Trial Balance.

5. Do you agree or disagree with the following statements.

Question 1.
A temporary account opened to rectify the entry is known as suspense A/c.
Answer:
Agree

Question 2.
Rectified entries are passed in Ledger.
Answer:
Disagree

Question 3.
Compensating errors affect the agreement of Trial Balance.
Answer:
Disagree

Question 4.
There is only one type of error.
Answer:
Disagree

Question 5.
Transaction recorded without following the accounting principles and rules are known as Errors of Principle.
Answer:
Agree

6. Complete the following sentences.

Question 1.
___________ sided errors affect the total of Trial Balance.
Answer:
One

Question 2.
___________ sided errors do not affect the Trial Balance.
Answer:
Two

Question 3.
One sided error do not require ___________ entry.
Answer:
Rectification

Question 4.
Errors which are committed in writing the accounts are called error of ___________
Answer:
Posting

Question 5.
Under casting of Sales book is corrected by ___________ Sales Account.
Answer:
Crediting

Question 6.
The disagreement of Trial Balance indicates that an ___________ has been committed.
Answer:
Error

Question 7.
An asset has been purchased for the firm. However, the amount was debited to the purchase account. It is an error of ___________
Answer:
Principle

Question 8.
___________ account is necessary for rectification of one-sided error.
Answer:
Suspense A/c

Question 9.
An item of ₹ 95 has been debited to a personal account as ₹ 59. It is an error of ___________
Answer:
Commission

Question 10.
Rectification entries are passed in ___________
Answer:
Journal Proper

Question 11.
Two sided errors are rectified by passing ___________ entry.
Answer:
Rectification

Question 12.
If the Trial Balance does not tally, its difference is transferred to ___________ Account.
Answer:
Suspense

Question 13.
Under casting of an account is ___________ sided error.
Answer:
One

Question 14.
If the transaction is not at all recorded in the books, it is called an error of ___________
Answer:
Complete Omission

Question 15.
If the total balance shows short a credit, the suspense account will have a ___________ balance.
Answer:
Credit

Question 16.
An error that affects debit as well as credit side, it is called as ___________ errors.
Answer:
Two-Sided

Question 17.
Wages paid for installation of machinery debited to Wages Account is an error of ___________
Answer:
Principle

Question 18.
Errors which cancel out the effect of one another is called ___________ errors.
Answer:
Compensating

Question 19.
One sided errors are disclosed by ___________
Answer:
Trial Balance

Question 20.
In an error of omission, the debit and credit are ___________
Answer:
Equal

Practical Problems

Question 1.
The trial balance of Sagar did not agree. It showed an excess credit of ₹ 7,550. Sagar put the difference to the suspense account. He located the following errors:
1. Sales return book was overcast by ₹ 1,200.
2. Purchase book was undercast by ₹ 750.
3. Goods returned to Mahesh ₹ 1,000 were recorded through the sales book.
4. Credit purchases from Mahadev ₹ 6,000 were recorded through the sales book.
5. Credit purchases from Damodhar ₹ 4,000 were recorded through the sales book. However, Damodhar’s account was correctly credited.
6. Salary paid ₹ 3,500 was debited to the employee’s personal account.
Give journal entries to rectify the above errors and prepare Suspense Account.
Solution:
Journal Proper

Question 2.
The trial balance of Radhika did not agree. Radhika put the difference to the suspense account. Subsequently, she located the following errors.
1. Furniture purchased for ₹ 6,000 was posted to the purchases account as ₹ 600.
2. Repairs to Machinery ₹ 500 were debited to the Machinery account.
3. Wages paid for the installation of Machinery ₹ 750 was posted to wages account.
4. Purchased material ₹ 8,000 and Wages ₹ 2,000 were used for construction of the building. No adjustment was made in the books.
5. Total of sales returns book ₹ 2,000 was not posted to the ledger.
6. Old Furniture sold to Dinesh at its book value of ₹ 2,500 was recorded through sales book.
Give journal entries and prepare Suspense Account.
Solution:
Journal Proper

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 3 Basic Analytical Techniques Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 3 Basic Analytical Techniques

Question 1.
Give reason: Purification of a chemical substance is important before investigating its composition and properties.
Answer:

• Chemical substances occur in nature in impure stage.
• Also, chemical substances synthesized in the laboratory are obtained in crude and impure form.
• Impurities present in the chemical substances may interfere with the properties to be determined (e.g. melting point or boiling point).
• Therefore, before investigating composition and properties of a given chemical substance, it is important to obtain it in the pure form.

Question 2.
What are the different types of impurities that a solid may contain?
Answer:
A solid substance may contain two types of impurities:

• Impurities which are soluble in the same solvent as the main substance.
• Impurities which are not soluble in the same solvent as the main substance.

Question 3.
For which of the following cases, is the process of filtration feasible? Why?
Case 1: A solid substance containing impurities which are soluble in the same solvent as the main substance.
Case 2: A solid substance containing impurities which are not soluble in the same solvent as the main substance.
Answer:
Impurities which are not soluble in the same solvent as the main compound can be separated by a simple process called filtration. Hence, for ‘Case 2’, filtration is more feasible.

Question 4.
Describe the process of filtration with a neat and labelled diagram.
Answer:
i. Impurities which are not soluble in the same solvent as the main compound can be separated by a simple process called filtration.
ii. Procedure:
a. A circular piece of filter paper is folded to form a cone and fitted in the funnel.
b. The funnel is fixed on a stand and a beaker is kept below.
c. The mixture which has to be purified is added to a suitable solvent in which the main compound dissolves.
d. The paper is made moist, and the solution to be filtered is poured on the filter paper.
e. Diagram:

iii. The insoluble part remaining on the filter paper is called residue and the liquid which pass through the filter paper and collected in the beaker is called filtrate.
iv. This process is similar to separating tea leaves from decoction of tea or sand from mixture of sand and water.

Question 5.
Why is safety bottle used when filtration is carried out under suction?
Answer:
The safety bottle is used to prevent sucking of the filtrate into suction pump.

Question 6.
Name the steps involved in the process of crystallization.
Answer:
Steps involved in the process of crystallization:

• Preparation of a saturated solution
• Hot filtration
• Cooling of the filtrate
• Filtration

Question 7.
How is saturated solution of the crude solid prepared?
Answer:

• A saturated solution of the crude solid is prepared by boiling it in a small but sufficient quantity of a suitable solvent.
• The main solute from the sample of the crude solid dissolves to form a saturated solution on boiling.

[Note: The solution is not saturated with respect to the soluble impurities, as they are in small proportion.]

Question 8.
Explain the following steps with respect to the process of crystallization.
i. Preparation of a saturated solution
ii. Hot filtration
iii. Cooling of the filtrate
iv. Filtration
Answer:
i. Preparation of a saturated solution:

• A saturated solution of the crude solid is prepared by boiling it in a small but sufficient quantity of a suitable solvent.
• On doing so the main solute forms an almost saturated solution, but the solution is not saturated with respect to the soluble impurities, as they are in small proportion.

ii. Hot filtration: The hot saturated solution is quickly filtered to remove undissolved impurities as residue. Filtration under suction can be employed for rapid filtration.

iii. Cooling of the filtrate:

• The hot filtrate is allowed to cool.
• On cooling, the filtrate becomes supersaturated with respect to the main dissolved solute because solubility of a substance decreases with lowering of temperature.
• The excess quantity of the dissolved solute comes out of the solution in the form of crystals.
• The dissolved impurities, however, do not supersaturate the solution, as their quantity is small.
• These continue to stay in the solution in dissolved state even on cooling. Therefore, the separated crystals are free from soluble impurities.

iv. Filtration:

• The crystals obtained on cooling are further purified by filtration to remove insoluble impurities.
• The filtrate obtained is called as mother liquor.
• The crystals obtained after filtration are free from soluble as well as insoluble impurities.

Question 9.
Name the common solvents used in the process of crystallization.
Answer:
The commonly used solvents are water, ethyl alcohol, methyl alcohol, acetone, ether or their combinations.

Question 10.
Describe the process of crystallization of common salt from impure sample with the help of a diagram.
Answer:

• Impure sample of a common salt is added to the required quantity of water and stirred with a glass rod.
• More amount of salt is added and the solution is heated till no more salt dissolves.
• The hot saturated solution is filtered off to remove insoluble impurities while the filtrate is collected in an evaporating dish.
• The filtrate is allowed to cool which results in the formation crystals of pure salt (NaCl) leaving behind the soluble impurities.
• The crystals are filtered and dried.

The diagram is as follows:

Question 11.
Which solvent is used for the purification of copper sulphate and benzoic acid?
Answer:
The solvent used for the purification of copper sulphate and benzoic acid is water.

Question 12.
Define: Fractional crystallization
Answer:
Fractional crystallization is a process wherein two or more soluble substances having widely different solubilities in the same solvent at same temperature are separated by crystallization.

Question 13.
Give a brief description of the principle of fractional crystallization.
Answer:
Fractional crystallization is based on the differences in solubilities of two or more compounds in the same solvent at the same temperature. That is, the substance which is least soluble crystallizes out first and the most soluble substance crystallizes out last.
e.g. Mixture of two solutes A and B can be purified by fractional crystallization as follows:

• Preparation of a saturated solution: Mixture of two solutes A and B are dissolved in a suitable hot solvent to prepare a saturated solution.
• Hot filtration: The hot saturated solution is filtered to remove insoluble impurities.
• Cooling of the filtrate: Hot filtrate is allowed to cool. On cooling, the solute which is least soluble crystallizes out first leaving behind the most soluble substance in the mother liquor.
• Filtration: The crystals formed are filtered, washed with solvent and dried. Crystals obtained will be of a solute which is least soluble in a given solvent.
• Concentration of a mother liquor: The mother liquor is concentrated by evaporating the solvent. These crystals are filtered and dried to obtain the second purified component (which was more soluble in given solvent).

Question 14.
Which type of impure liquids can be purified by the process of distillation?
Answer:
Distillation technique can be employed for the purification of

• volatile liquids from non-volatile impurities.
• liquids having sufficient difference in their boiling point.

Question 15.
Explain the construction of simple distillation unit using neat labelled diagram.
Answer:
i. The apparatus used for simple distillation is shown in the figure below:

ii. It consists of round bottom flask fitted with a cork having a thermometer.
iii. The flask has a sidearm through which it is connected to a condenser.
iv. The condenser has a jacket with two outlets through which water is circulated.
v. The liquid to be distilled is taken in the round bottom flask fixed by clamp.
vi. The flask is placed in a water bath or oil bath or sometimes wire gauze is kept on a stand as shown in the figure.

Question 16.
State the principle involved and describe the process to separate acetone and water from their mixture.
Answer:
i. Acetone and water can be separated from their mixture by simple distillation.

ii. Principle: Acetone and water are two miscible liquids having an appreciable difference (more than 30 K) in their boiling points. Acetone boils at 56 °C while boiling point of water is 100 °C. When the mixture of acetone and water is heated and temperature of the mixture reaches 56 °C acetone will distil out first. Once all acetone distils out, and when the temperature rises to 100 °C water will distil out.

iii. Process to separate acetone and water from their mixture:

• Take the mixture of water and acetone in the distillation flask.
• Heat the flask on a water bath carefully. At 56 °C acetone will distil out, collect it in receiver.
• After all acetone distilled, change the receiver. Discard a few mL of the liquid. As the temperature reaches 100 °C water will begin to distil. Collect this in another receiver.

Question 17.
What is the advantage of fractional distillation over simple distillation?
Answer:
If in a mixture, the difference in boiling points of two liquids is not appreciable/large, they cannot be separated using simple distillation. To separate such liquids, fractional distillation is used.

Question 18.
Label the following diagram and explain the process by giving example.

Answer:
The labelled diagram is as follows:

i. In fractional distillation, vapours first pass through the fractionating column.
ii. Vapours of more volatile liquid with lower boiling point rise up more than the vapours of liquid having higher boiling point.
e.g.

• Suppose we have a mixture of two liquid ‘A’ and ‘B’ having boiling points 363 K and 373 K respectively.
• ‘A’ is more volatile and ‘B’ is less volatile. As the mixture is heated, vapours of ‘A’ along with a little vapours of ‘B’ rise up and come in contact with the large surface of the fractionating column.
• Vapours of ‘B’ condense rapidly into the distillation flask. While passing through the fractionating column, there is an exchange between the ascending vapours and descending liquid. The vapours of ‘B’ are scrubbed off by the descending liquid, this makes the vapours richer in ‘A’.
• This process is repeated each time the vapours and liquid come in contact with the surface in the fractionating column.
• Rising vapours become richer in ‘A’ and escape through the fractionating column and reach the condenser while the liquid in the distillation flask is richer in ‘B.
• The separated components are further purified by repeating the process.

Question 19.
Give two examples of a mixture that can be separated by fractional distillation.
Answer:

1. Mixture of acetone (b.p. 329 K) and methyl alcohol (b.p. 337.7 K)
2. Mixture of acetone (b.p. 329 K) and benzene (b.p. 353 K)

Question 20.
Give one industrial application of fractional distillation.
Answer:
Fractional distillation is used in petroleum industry to separate different fractions of crude oil.

Question 21.
Write a short note on distillation under reduced pressure.
Answer:

• Liquids having very high boiling points or which decompose on heating are purified by the method of distillation under reduced pressure.
• In this method, the liquid is made to boil at a temperature lower than its normal boiling point by reducing the pressure on its surface.
• The external pressure is reduced using a water pump or vacuum pump, e.g. Glycerol can be separated from soap by using this method.

Question 22.
Write the principle of solvent extraction and explain the process with labelled diagram.
Answer:
Principle: Extraction of compound takes place based on the difference in solubility of compound in two liquids.

• In this process, the solute distributes itself between two immiscible liquids. From the aqueous phase the solute gets extracted in the organic phase.
• On shaking for a few times with small volumes of organic phase, most of the solute gets extracted into the organic phase.
• Then solute is then recovered from organic solvent either by evaporation of organic solvent or distillation.

Diagram:

Question 23.
Write a short note on continuous extraction method.
Answer:

• During solvent extraction, if the solute is found to be less soluble in organic phase, then continuous extraction method is employed.
• In this method, the same amount of organic solvent is used repeatedly for extraction.
• This ensures that the most of the solute gets extracted in the organic phase.
• This technique involves continuous distillation of the solvent within the same assembly. Hence, the use of large quantity of organic solvent is avoided.

Question 24.
Match the following:

	Process		Used in the purification/separation of
i.	Crystallization	a.	Acetone and benzene
ii.	Simple distillation	b.	Benzoic acid and water
iii.	Fractional distillation	c.	Impure copper sulphate
iv.	Solvent extraction	d.	Acetone and water

Answer:
i – c,
ii – d,
iii – a,
iv – b

Question 25.
What is chromatography? Explain the principle behind it.
Answer:
Chromatography is a technique used to separate components of a mixture, and also purify compounds.
Principle: The principle of separation of substances in chromatography is based on the distribution of the solutes in two phases, i.e., stationary phase and mobile phase.

• Chromatography uses two phases for separation.
• This technique is based on the difference in rates at which components in the mixture move through the stationary phase under the influence of the mobile phase.
• In this technique, first the mixture of components is loaded at one end of the stationary phase and then the mobile phase is allowed to move over the stationary phase. The mobile phase can be a pure solvent or a mixture of solvents.
• Depending on the relative affinity of the components toward the stationary phase and mobile phase, they remain on the surface of the stationary phase or move along with the mobile phase, and gradually get separated.

Question 26.
Give a brief description of column chromatography with an illustration.
Answer:
Column chromatography involves the separation of components over a column of stationary phase. The stationary phase material can be alumina, silica gel.
Procedure:

• A slurry of the stationary phase material is filled in a long glass tube provided with a stopcock at the bottom and a glass wool plug at the lower end.
• The mixture to be separated is dissolved in a suitable solvent and then it is loaded on top of adsorbent column.
• A suitable mobile phase which could be a single solvent or a mixture of solvents is then poured over the adsorbent column.
• The mixture along with the mobile phase slowly moves down the column.
• The solutes get adsorbed on the stationary phase and depending on the degree to which they are adsorbed, they get separated from each other.
• The component which is readily adsorbed are retained on the column and others move down the column to various distances forming distinct bands.
• The component which is less strongly adsorbed is desorbed first and leaves the column first, while the strongly adsorbed component is eluted later.
• The solutions of these components are collected separately.
• These different components can be recovered by evaporating the solvent.

Question 27.
How is TLC plate or chromplate prepared?
Answer:
TLC plate or chromplate is prepared by applying a thin layer (0.2 mm thick) of adsorbent silica gel or alumina spread over a glass plate.

Question 28.
Describe the process of thin layer chromatography (TLC) and separation of components in it.
Answer:
i. Process:

• A thin layer (about 0.2 mm thick) of an adsorbent like silica gel or alumina is spread over a thin glass plate (called chromplate or TLC plate). This plate acts as a stationary phase.
• With the help of a capillary tube, the solution of the mixture to be separated is spotted at above 2 cm (on base line) from one end of the TLC plate.
• The TLC plate is then placed in a closed jar containing a suitable solvent (mobile phase or eluant).
• As the mobile phase rises up the plate, the components of the mixture move up along with the mobile phase to different distances depending upon their degree of adsorption, thus resulting in complete separation.

ii. Separation of components:

• If the components are coloured, they appear as separated coloured spots on the plate.
• If the components are not coloured but have property of fluorescence, they can be visualised under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The Iodine vapours are adsorbed by the components and the spots appear brown.
• Amino acids are visualised by spraying the plate with a solution of ninhydrin. This is known as spraying agent.

Question 29.
Name the physical state each of stationary phase and mobile phase in partition chromatography.
Answer:
In partition chromatography, both stationary and mobile phases are in liquid state.

Question 30.
State the principle of partition chromatography.
Answer:
Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.

Question 31.
Describe the process of paper chromatography.
Answer:
Process of paper chromatography:

• The mixture of the compound to be analysed is dissolved in a suitable solvent and spotted on the chromatography paper about 2 cm from one end of the paper using a glass capillary.
• The paper is then suspended in a chamber containing the mobile phase.
• The mobile phase rises up the paper and flows over the spot, due to capillary action.
• Different solutes are retained differently on the paper depending on their selective partitioning between the two phases. The paper strip so developed, is known as chromatogram.

Question 32.
Name the following:
i. A glass plate coated with a thin layer of silica gel.
ii. A spraying agent used for the visualization of amino acids.
Answer:
i. Chromplate/TLC plate
ii. Ninhydrin

Question 33.
Write a short note on R_f value.
Answer:
i. In chromatography, migration of the solute relative to the solvent front gives an idea about the relative retention of the solutes (or components of t the mixture) on the stationary phase.
ii. The relative adsorption of solutes is expressed in terms of its R_f value.
The symbol R_f stands for Retardation Factor.

Question 34.
In a chemical laboratory, Priyal was asked to isolate an organic compound from its aqueous solution. She added ethyl acetate to the given sample, separated the organic layer and kept it for evaporation. At the end of her practical, Priyal found few crystals in the beaker which she kept for evaporation. Answer the following questions:
i. In the above passage, which method was used by Priyal for separation? State its principle.
ii. Why do you think the organic compound dissolved in ethyl acetate?
iii. Illustrate the method of separation used in the passage with an example.
Answer:
i. Method used: Solvent extraction method.
Principle: Extraction of compound takes place based on the difference in solubility of compound in two liquids,

• In this process, the solute distributes itself between two immiscible liquids. From the aqueous phase the solute gets extracted in the organic phase.
• On shaking for a few times with small volumes of organic phase, most of the solute gets extracted into the organic phase.
• Then solute is then recovered from organic solvent either by evaporation of organic solvent or distillation.

ii. An organic compound (non-polar) dissolves in organic solvents (non-polar) because of the dipole-dipole interactions in between them (like dissolves like). Water is a polar solvent and it is unlikely that the covalent constituents of the organic substance is strong enough to break the ionic bonds. Any substance dissolves in other because it is able to break the bonds between the solvent molecules and form weak bonds with the solvent molecules. Hence, the organic compound will be more soluble in ethyl acetate as compared to water and this helps in its isolation from aqueous solution.

iii. An example for the separation of organic compound using solvents extraction method is: Benzoic acid in water can be extracted from its aqueous solution by using benzene.

Question 35.

Based on the above diagram, answer the following questions:
i. Name the chromatographic technique involved.
ii. From the developed chromatogram, state which has the highest and which has the lowest R_f value?
iii. Based on the TLC, which component would elute out at the end of a column chromatography?
iv. Mention two applications of TLC method.
Answer:
i. Thin layer chromatography
ii. Based on the developed chromatogram, spot ‘x’ has the highest Rf value while spot ‘z’ the lowest Rf value.
iii. Based on the TLC, spot ‘z’ being strongly adsorbed will elute at the end of a column chromatography.
iv. Applications of TLC are:

• Separation of plant pigments from its mixture.
• Separation of impurities from a given organic compound.
• Separation of different amino acid.

Multiple Choice Questions

1. If a crude solid is made of mainly one substance and has some impurities then it is purified by ……………..
(A) crystallization
(B) distillation
(C) extraction
(D) sublimation
Answer:
(A) crystallization

2. Impure common salt can be purified by ……………
(A) crystallization
(B) distillation
(C) extraction
(D) sublimation
Answer:
(A) crystallization

3. Which of the following solvents is most commonly used for the crystallization of copper sulphate?
(A) Water
(B) Acetone
(C) Ether
(D) Methanol
Answer:
(A) Water

4. In distillation of liquid, water condenser is used ……………
(A) to boil the liquid
(B) to collect the liquid
(C) to condense hot vapours of the liquid
(D) to adsorb the liquid
Answer:
(C) to condense hot vapours of the liquid

5. Separation of binary mixture of acetone and methyl alcohol is done by ……………
(A) simple distillation
(B) fractional distillation
(C) fractional crystallization
(D) re-crystallization
Answer:
(B) fractional distillation

6. Which of the following method is used to separate different fractions of crude oil?
(A) Solvent extraction
(B) Simple distillation
(C) Fractional distillation
(D) TLC
Answer:
(C) Fractional distillation

7. The method used to separate a given organic compound present in aqueous solution by shaking with a suitable solvent in which the compound is more soluble than water is called ……………….
(A) simple distillation
(B) fractional distillation
(C) solvent extraction
(D) crystallization
Answer:
(C) solvent extraction

8. Adsorption chromatography is a chromatographic technique based on the principle of ……………
(A) differential adsorption
(B) differential solubility
(C) differential extraction
(D) all of these
Answer:
(A) differential adsorption

9. The stationary phase and mobile phase in TLC are ……………. respectively.
(A) solid and liquid
(B) solid and gas
(C) liquid and solid
(D) liquid and liquid
Answer:
(A) solid and liquid

10. Which of.the following is most commonly used for the visualization of amino acids in chromatography?
(A) Ultraviolet light
(B) Spraying agent
(C) Sunlight
(D) X-rays
Answer:
(B) Spraying agent

11. The stationary phase and mobile phase in partition chromatography are ………….. respectively.
(A) solid and liquid
(B) solid and gas
(C) liquid and solid
(D) liquid and liquid
Answer:
(D) liquid and liquid

12. Paper chromatography is based on the principle of …………….
(A) adsorption
(B) partition
(C) solubility
(D) volatility
Answer:
(B) partition

13. In paper chromatography, the mobile phase rises up the chromatography paper due to ………………
(A) evaporation of volatile solvent
(B) capillary action
(C) gravitational force
(D) differential adsorption
Answer:
(B) capillary action

14. Which of the following is a type of partition chromatography?
(A) Column chromatography
(B) Thin layer chromatography
(C) Paper chromatography
(D) Both (B) and (C)
Answer:
(C) Paper chromatography

15. The principle of differential adsorption is applicable for which of the following chromatographic technique?
(A) Column chromatography
(B) Thin layer chromatography
(C) Paper chromatography
(D) Both (A) and (B)
Answer:
(D) Both (A) and (B)

16. Which of the following method will give clean separation of sample of chloroform (organic liquid) and water in short time span?
(A) TLC
(B) Distillation under reduced pressure
(C) Solvent extraction
(D) Simple distillation
Answer:
(C) Solvent extraction

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 14 Ecosystems and Energy Flow Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Multiple choice questions

Question 1.
What is a true about ecosystem?
(a) Primary consumers are least dependent upon producers.
(b) Primary consumers outnumber the producers.
(c) Producers are more than primary consumers.
(d) Secondary consumers are the largest and most powerful.
Answer:
(c) Producers are more than primary consumers.

Question 2.
In an ecosystem, which shows one way passage?
(a) Free energy
(b) Carbon
(c) Nitrogen
(d) Potassium
Answer:
(a) Free energy

Question 3.
How many are biotic components from the following? Climate, carbohydrates, microbes, green plants, lipids, water, proteins, photosynthetic bacteria, chemosynthetic bacteria, herbivores, carnivores.
(a) 6
(b) 5
(c) 4
(d) 7
Answer:
(a) 6

Question 4.
From the following, which is the basic requirement for any type of ecosystem to function and sustain?
(a) Constant output of solar energy
(b) Constant input of solar energy
(c) Organic substances
(d) Organic substances dissolved in water
Answer:
(b) Constant input of solar energy

Question 5.
Which type of spatial patterns noticed in ecosystem structure?
(a) Zonation
(b) Stratification
(c) Zonation and stratification
(d) Zonation, stratification and distribution
Answer:
(c) Zonation and stratification

Question 6.
In which strata of any aquatic body there will be maximum photosynthesis?
(a) Bottom deposits
(b) Middle strata of a water body
(c) Near coastal region
(d) The depth till where sunlight can reach
Answer:
(d) The depth till where sunlight can reach

Question 7.
Which spatial pattern occurs vertically in an ecosystem?
(a) Zonation
(b) Stratification
(c) Y-shaped pattern
(d) Pyramid
Answer:
(b) Stratification

Question 8.
Which of the following is one of the characteristics of a biological community?
(a) Sex ratio
(b) Stratification
(c) Natality
(d) Mortality
Answer:
(b) Stratification

Question 9.
Identify the likely organisms (1), (2), (3) and (4) in the food web given below:

Answer:
(a) deer, rabbit, frog, rat

Question 10.
Which of the following is the basic requirement for any ecosystem to function and sustain?
(a) A constant input of solar energy
(b) Ample water availability
(c) Cool temperatures
(d) Enough winds and currents around
Answer:
(a) A constant input of solar energy

Question 11.
Which one is the most accurate definition of primary production?
(a) The amount of the food produced by plants.
(b) The amount of food available to herbivores.
(c) The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.
(d) The amount of crop produced by farmer.
Answer:
(c) The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.

Question 12.
Primary production is expressed in terms of
(a) kg/area
(b) weight (g^-2) or energy (kcal^{m -2})
(c) energy in food calories
(d) calories /sq. m.
Answer:
(b) weight (g^-2) or energy (kcal^m-2)

Question 13.
Considerable amount of GPP is utilized by plants in
(a) metabolism
(b) respiration
(c) homeostasis
(d) excretion
Answer:
(b) respiration

Question 14.
Which one is the most appropriate definition of secondary productivity?
(a) Secondary productivity is the rate of formation of new organic matter by consumers.
(b) Secondary productivity is the amount of food consumed by the carnivores.
(c) Secondary productivity is the amount of food consumed by the producers.
(d) Secondary productivity is the amount of energy lost in the food chain.
Answer:
(a) Secondary productivity is the rate of formation of new organic matter by consumers.

Question 15.
Choose the factor on which primary productivity does not depend.
(a) The plant species inhabiting a particular area
(b) Amount of primary consumers dependent on plants
(c) Availability of nutrients
(d) Photosynthetic capacity of plants
Answer:
(b) Amount of primary consumers dependent on plants.

Question 16.
The annual net primary productivity of the whole biosphere is approximately billion tonnes (dry weight) of organic matter.
(a) 10
(b) 50
(c) 100
(d) 170
Answer:
(d) 170

Question 17.
The rate at which light energy is changed into chemical energy of organic molecules in ecosystems is
(a) net primary productivity
(b) gross primary productivity
(c) net secondary productivity
(d) gross secondary productivity
Answer:
(b) gross primary productivity

Question 18.
What is net primary productivity?
(a) Total rate of photosynthesis
(b) Rate of energy storage by consumer
(c) Amount of organic matter stored by plants
(d) Rate of energy used
Answer:
(c) Amount of organic matter stored by plants

Question 19.
Which is the correct sequence of the steps decomposition process?
(a) Fragmentation → Mineralization → Catabolism → Humification → Leaching
(b) Leaching → Catabolism → Humification → Fragmentation → Mineralization
(c) Fragmentation → Leaching → Catabolism → Humification → Mineralization
(d) Catabolism → Humification → Fragmentation → Leaching → Mineralization
Answer:
(c) Fragmentation → Leaching → Catabolism → Humification → Mineralization

Question 20.
Match the columns:

1. Fragmentation	i. Release of inorganic nutrients
2. Leaching	ii. Bacterial and fungal enzymes
3. Catabolism	iii. Accumulation of dark amorphous substance
4. Humification	iv. Precipitated as unavailable salts
5. Mineralization	v. Break down into smaller pieces

(a) 1-v, 2-ii, 3-iii, 4-i, 5-iv
(b) 1-v, 2-iv, 3-ii, 4-iii, 5-i
(c) 1-ii, 2-iii, 3-iv, 4-v, 5-i
(d) 1-iii, 2-v, 3-iv, 4-i, 5-ii
Answer:
(b) 1-v, 2-iv, 3-ii, 4-iii, 5-i

Question 21.
The slow rate of decomposition of fallen logs in nature is due to their
(a) low moisture content
(b) poor nitrogen content
(c) anaerobic environment around them
(d) low cellulose content
Answer:
(a) low moisture content

Question 22.
Small amount of energy sustains the entire living world is only ……………………… of the PAR.
(a) 20-25%
(b) 10-15%
(c) 2-10%
(d) 1-2%
Answer:
(c) 2-10%

Question 23.
Choose incorrect statement out of the following
(a) Much larger fraction of energy flows through the DFC than through the GFC.
(b) Some of the organisms of DFC are prey to the GFC animals.
(c) In an aquatic ecosystem, GFC is the major conduit for energy flow.
(d) Much less fraction of energy flows through the GFC than through the DFC.
Answer:
(d) Much less fraction of energy flows through the GFC than through the DFC

Question 24.
In an ecosystem, bacteria are considered as
(a) primary consumers
(b) microconsumers
(c) macroconsumers
(d) secondary consumers
Answer:
(b) microconsumers

Question 25.
Which of the following can be a top carnivore of marine ecosystem?
(a) Zooplankton
(b) Sea cucumber
(c) Sea horse
(d) Kingfisher
Answer:
(d) Kingfisher

Question 26.
Identify the possible link ‘A’ in the following food chain.
Plant → Insect → Frog → A’ → Eagle
(a) Rabbit
(b) Wolf
(c) Cobra
(d) Parrot
Answer:
(c) Cobra

Question 27.
Why is the pyramid of biomass inverted in sea?
(a) Because plants are absent.
(b) Because fishes have less biomass.
(c) Because phytoplankton which are primary producers have less biomass.
(d) Because primary producers have more biomass.
Answer:
(c) Because phytoplankton which are primary producers have less biomass.

Question 28.
Identify the wrong statement in relation to concept of pyramids.
(a) Pyramid of energy is always upright, can never be inverted.
(b) In most ecosystems, all the pyramids of number of energy and biomass are upright.
(c) Inverted pyramid of biomass is observed in pond ecosystem as small standing crop of phytoplankton supports larger zooplanktons.
(d) Pyramid of biomass in sea is upright because fishes feed on standing crop of planktons.
Answer:
(d) Pyramid of biomass in sea is upright because fishes feed on standing crop of planktons.

Question 29.
Identify A, B, C and D in the following simplified model of phosphorus cycling in a terrestrial ecosystem.

Answer:
(a) Detritus, Rock minerals, producer, litter fall

Question 30.
From the following, which is not a fate of carbon in plants?
(a) Released in transpiration. atmosphere through
(b) Liberated in respiration. atmosphere through
(c) Consumed by animal in the form of food.
(d) Remains as it is as organic matter when plant dies.
Answer:
(a) Released in atmosphere through transpiration.

Question 31.
Bacterial role in carbon cycle is ………………
(a) photosynthesis
(b) chemosynthesis
(c) assimilation
(d) breakdown of organic matter
Answer:
(d) breakdown of organic matter

Question 32.
Which out of the following is the major reservoir of carbon?
(a) Animal bodies
(b) Fruits
(c) Ocean
(d) Coal mines
Answer:
(c) Ocean

Question 33.
Which act of human beings has great impact on carbon cycle?
(a) Breaking of limestone
(b) Building up of limestone reefs
(c) Burning of fossil fuels
(d) Volcanic eruption
Answer:
(c) Burning of fossil fuels

Question 34.
Rock phosphates are brought into circulation by the process of ………………….
(a) combustion
(b) sedimentation
(c) weathering
(d) absorption
Answer:
(c) weathering

Question 35.
In animals this structure is not containing phosphorus.
(a) Bones
(b) Shells
(c) Teeth
(d) Hairs
Answer:
(d) Hairs

Question 36.
Phosphorus is not a major constituent of
(a) DNA
(b) Proteins
(c) RNA
(d) ATP
Answer:
(b) Proteins

Question 37.
Secondary Succession takes place on/in
(a) Newly cooled lava
(b) Bare rock
(c) Degraded forest
(d) Newly created pond
Answer:
(c) Degraded forest

Question 38.
Vertical distribution of different species occupying different levels in a biotic community is known as
(a) Pyramid
(b) Divergence
(c) Stratification
(d) Zonation
Answer:
(c) Stratification

Question 39.
The developmental stages of ecological succession are called
(a) serial stages
(b) serai stages
(c) cereal stages
(d) trophic levels
Answer:
(b) serai stages

Question 40.
What is the peculiarity of pioneers in succession?
(a) They are always heterotrophic components.
(b) They always face favourable growth conditions.
(c) They are the most successful terminal occupants of the area.
(d) They face adverse conditions and get established there.
Answer:
(d) They face adverse conditions and get established there.

Question 41.
What are pioneers?
(a) First serai stage
(b) Heterotrophic serai stage
(c) Terminal (last) serai stage
(d) Final climax community
Answer:
(a) First serai stage

Question 42.
The stable community established in an area is known as
(a) pioneer community
(b) climax community
(c) equilibrium community
(d) autotrophic community
Answer:
(b) climax community

Question 43.
The entire sequence of communities that successively change in a given area are called
(a) climax
(b) sere
(c) pioneer
(d) standing population
Answer:
(b) sere

Question 44.
The primary succession refers to the development of communities on a ………………….
(a) freshly cleared crop field
(b) forest clearing after devastating fire
(c) pond, freshly filled with water after a dry phase
(d) Newly-exposed habitat with no record of earlier vegetation
Answer:
(d) Newly-exposed habitat with no record of earlier vegetation

Question 45.
Which is the most important service provided by environment?
(a) Release of oxygen
(b) Formation of ozone layer
(c) Carbon assimilation in photosynthesis
(d) Agents of pollination
Answer:
(c) Carbon assimilation in photosynthesis

Match the columns

Question 1.

Column A	Column B
(1) Tansley	(a) 10% law
(2) C. Elton	(b) Ecosystem services
(3) R. Lindemann	(c) Ecological pyramids
(4) Millennium Ecosystem Assessment report	(d) Ecosystem

Answer:

Column A	Column B
(1) Tansley	(d) Ecosystem
(2) C. Elton	(b) Ecosystem services
(3) R. Lindemann	(a) 10% law
(4) Millennium Ecosystem Assessment report	(c) Ecological pyramids

Question 2.

Column A	Column B
(1) Rooted floating angiosperm	(a) Cyperus
(2) Free-floating plant	(b) Typha
(3) Reed swamp	(c) Pistia
(4) Marsh-meadow	(d) Lotus

Answer:

Column A	Column B
(1) Rooted floating angiosperm	(d) Lotus
(2) Free-floating plant	(c) Pistia
(3) Reed swamp	(b) Typha
(4) Marsh-meadow	(a) Cyperus

Question 3.

Column A	Column B
(1) Pioneer species	(a) Entire gradient of communities
(2) Climax species	(b) Spatial pattern
(3) Succession	(c) Quercus
(4) Sere	(d) Crustose lichen

Answer:

Column A	Column B
(1) Pioneer species	(d) Crustose lichen
(2) Climax species	(c) Quercus
(3) Succession	(b) Spatial pattern
(4) Sere	(a) Entire gradient of communities

Classify the following to form Column B as per the category given in Column A.

Question 1.
Estuarine waters, Taiga, Aquarium tank, Lake, Evergreen forest, Flower garden.

Types of ecosystems	Examples
(1) Aquatic	————–
(2) Terrestrial	—————
(3) Artificial	————–

Answer:

Types of ecosystems	Examples
(1) Aquatic	(a) Estuarine waters, Lake
(2) Terrestrial	(b) Taiga, Evergreen forest
(3) Artificial	(c) Aquarium tank, Flower garden

Question 2.
Xerarch, Epipelagic, Hydrosere, Sublittotral, Intertidal, Benthic.

Phenomena	Examples
(1) Stratification	————–
(2) Zonation	—————
(3) Succession	————–

Answer:

Phenomena	Examples
(1) Stratification	(a) Epipelagic, Benthic
(2) Zonation	(b) Sublittotral, Intertidal
(3) Succession	(c) Xerarch, Hydrosere

Question 3.
Nature trails, Pollination, Sea food, Carbon sequestration, Animal therapy, Pest control, Health care, Nutrient cycling.

Column A	Column B
(1) Supporting services	————–
(2) Provisioning services	—————
(3) Regulating services	————–
(4) Cultured services	—————

Answer:

Column A	Column B
(1) Supporting services	(a) Pollination, Nutrient services cycling
(2) Provisioning services	(b) Sea food, Health care
(3) Regulating services	(c) Carbon sequestration, services Pest control
(4) Cultured services	(d) Nature trails, Animal services therapy

Very short answer questions

Question 1.
What forms the physical structure of the ecosystems?
Answer:
Interaction of biotic and abiotic components, results in a physical structure of ecosystems.

Question 2.
How is species composition of an ecosystem decided?
Answer:
By identification and enumeration of plant and animal species of a given ecosystem, its species composition can be decided.

Question 3.
What does base of each pyramid represent?
Answer:
The base of each pyramid represents the first trophic level of producers.

Question 4.
How is stratification seen in the forested land?
Answer:
Stratification seen in forested land is as follows trees occupying top vertical strata or layer of a forest, shrubs the second strata and herbs and grasses occupying the bottom layer.

Question 5.
What is meant by nutrient cycling?
Answer:
The cyclic movement of nutrient elements through the various components of an ecosystem, is called nutrient cycling.

Question 6.
Why nutrient cycling is called biogeochemical cycle?
Answer:
The nutrients are cycled from biotic organisms (bio) to abiotic components in the earth (geo) and all these nutrients are in the form of chemicals, therefore, nutrient cycling is called biogeochemical cycle.

Question 7.
What is the function of reservoirs in nutrient cycling?
Answer:
The function of the reservoir is to meet with the deficit, which occurs due to imbalance in the rate of influx and efflux in any ecosystem.

Question 8.
Till what time does the climax community remain stable?
Answer:
The climax community remains stable as long as the environment remains unchanged.

Question 9.
Which are the pioneers in the aquatic habitat during primary succession?
Answer:
Small phytoplankton are the pioneers in the aquatic habitat during primary succession.

Question 10.
What are serai communities?
Answer:
The individual transitional communities formed during succession are termed serai communities.

Question 11.
What are the benefits given by the healthy ecosystems?
Answer:
Healthy ecosystems give the benefits from wide range of economic, environmental and aesthetic goods and services.

Question 12.
Which are the major producers in a terrestrial ecosystem?
Answer:
The major producers in a terrestrial ecosystem are herbaceous and woody plants.

Question 13.
Which are the major producers in an aquatic ecosystem?
Answer:
Major producers in an aquatic ecosystem are phytoplankton and algae.

Question 14.
Which event is the beginning of the detritus food chain or web?
Answer:
Death of any organism is the beginning of the detritus food chain or web.

Question 15.
State the 10% law. Who gave this law?
Answer:
R. Lindermann gave the 10% law which states that only 10% of the energy is transferred to each trophic level as net energy, from the previous trophic level. B

Give definitions of the following

Question 1.
Ecosystem
Answer:
A self-regulatory and self-sustaining structural and functional unit of biosphere in which there are interacting biotic and abiotic components is called an ecosystem.

Question 2.
Stratification
Answer:
Vertical distribution of different species of plants and animals occupying different levels, is known as stratification.

Question 3.
Zonation
Answer:
Horizontal distribution of plants and animals on land or in water, is called zonation.

Question 4.
Productivity
Answer:
Conversion of inorganic chemicals into organic material with the help of the radiant energy of the sun by the autotrophs and consumption of the autotrophs by heterotrophs.

Question 5.
Decomposition
Answer:
Decomposition is the break-down of dead organic material and mineralization of the dead matter.

Question 6.
Energy flow
Answer:
Energy flow is the unidirectional flow of energy from producers to consumers and finally dissipation and loss as heat.

Question 7.
Saprotrophs
Answer:
Organisms which can degrade the detritus and obtain their energy and nutrient requirements are called saprotrophs

Question 8.
Trophic level
Answer:
A specific place occupied by the organisms in the food chain is called their trophic level.

Question 9.
Ecological pyramid
Answer:
Ecological pyramid is the graphic representation showing relationship between the organisms of different successive trophic levels with respect to energy, biomass and number.

Question 10.
Leaching
Answer:
The precipitation of water-soluble inorganic nutrients into the soil horizon in the form of salts is called leaching.

Question 11.
Humification
The formation of humus is humification.

Question 12.
Mineralization
Answer:
The process in which humus is degraded by some microbes to release inorganic nutrients is called mineralization.

Question 13.
Eutrophication
Answer:
Eutrophication is the sudden influx of phosphorus in water bodies due to agricultural runoff or industrial effluents which are rich in phosphate content.

Question 14.
Ecological succession
Answer:
The gradual and predictable change in the species composition of a given area is called ecological succession.

Question 15.
Sere
Answer:
The entire sequence of communities that successively change in a given area is called a sere.

Name the following

Question 1.
Functional aspects of ecosystem.
Answer:

1. Productivity
2. Decomposition
3. Nutrient cycling
4. Energy flow.

Question 2.
Two types of spatial patterns.
Answer:

1. Stratification
2. Zonation.

Question 3.
Stratification seen in open seas.
Answer:
Epipelagic, mesopelagic, bathy-pelagic and benthic zones.

Question 4.
Zonation seen at the junction of aquatic and terrestrial ecosystems.
Answer:
Inter-tidal, Littoral, Sub-littoral zones.

Question 5.
Common herbivores in terrestrial ecosystem.
Answer:
Insects (grasshopper, aphids), birds (parrot) and some mammals (sheep, cattle, goat, donkey).

Question 6.
Secondary consumers.
Answer:
All carnivorous animals such as frogs, lizards, birds of prey, hunting animals like tiger, lion, wolf, etc.

Question 7.
Three types of food chains.
Answer:

1. Grazing
2. Detritus
3. Arasitic.

Question 8.
Different trophic levels.
Answer:
Producer, herbivore, primary carnivore, secondary carnivore, tertiary carnivore and ultimate carnivore are different trophic levels.

Question 9.
Three types of ecological pyramids.
Answer:

1. Pyramid of biomass
2. Pyramid of numbers
3. Pyramid of energy

Question 10.
The important steps in the process of decomposition.
Answer:

1. Fragmentation
2. Leaching
3. Catabolism
4. Humification
5. Mineralization

Question 11.
Sequential steps in process of succession.
Answer:

1. Nudation
2. Invasion
3. Ecesis
4. Aggregation
5. Competition and co-action
6. Reaction and stabilization

Question 12.
Types of Nutrient cycles.
Answer:

1. Gaseous and
2. Sedimentary.

Question 13.
Reservoir for the sedimentary cycle.
Answer:
Earth’s crust.

Question 14.
Reservoir for gaseous cycles.
Answer:
Atmosphere.

Distinguish between the following

Question 1.
Natural ecosystem and artificial ecosystem.
Answer:

Natural ecosystem	Artificial ecosystem
1. Natural ecosystems are naturally formed.	1. Artificial ecosystems are man made.
2. There are no human inputs in natural ecosystems.	2. Artificial ecosystem is based on all human inputs.
3. Natural ecosystems are self-sustainable.	3. Artificial ecosystems are not self-sustainable.
4. In natural ecosystem, energy and materials are used and reused in cyclic manner. E.g. Ocean, forest, wetlands, estuary.	4. In artificial ecosystem, Energy and materials have to be given by human intervention which requires constant inputs. E.g. Farm land, aquaculture ponds, aquarium in the house.

Question 2.
Primary and secondary succession.
Answer:

Primary succession	Secondary succession
1. The primary succession starts in the area where no living organisms ever existed.	1. The secondary succession starts in an area which has lost all the living organisms once existed.
2. Areas where primary succession starts are bare rock, newly formed pond, newly cooled lava, etc.	2. Abandoned farm, cut or burnt forest, flooded land’, etc. are areas where secondary succession begins.
3. Primary succession is a very slow process.	3. Secondary succession is comparatively a faster process.

Question 3.
Carbon cycle and phosphorus cycle.
Answer:

Carbon cycle	Phosphorus cycle
1. Carbon cycle is a gaseous cycle.	1. Phosphorus cycle is a sedimentary cycle.
2. Carbon cycle has higher speed.	2. Phosphorus cycle is very slow.
3. There is respiratory release of CO2 in carbon cycle.	3. There is no respiratory release of phosphorus.
4. Atmospheric inputs of carbon through rainfall are larger.	4. Atmospheric inputs of phosphorus through rainfall are much smaller.
5. Exchanges of phosphorus between organism and environment are negligible in carbon cycle due to photosynthesis and respiration.	5. Exchanges of phosphorus between organism and environment are negligible in phosphorus cycle.

Given reasons

Question 1.
All animals are called consumers.
Answer:

1. All animals depend on plants for food either directly as in case of herbivorous animals or indirectly as in case of carnivorous animals.
2. Animals cannot perform photosynthesis as they are not autotrophic.
3. They are heterotrophic and hence all are called consumers.

Question 2.
Food chains do not exist in isolation, but are always interconnected to form food web.
Answer:

1. Food chains start from producers and end with consumers.
2. But beyond secondary carnivores, the amount of energy available is too less.
3. Thus, there is no tertiary carnivore that feeds exclusively on secondary carnivore.
4. The secondary carnivore, however, many times will feed on herbivores directly.
5. Therefore, food chains do not exist in isolation, but are always interconnected to form food web, so that the stability of ecosystem is maintained.

Question 3.
The pyramid of biomass in the sea is inverted.
Answer:

1. The food chain in the sea is dependent on producers i.e. phytoplankton.
2. The biomass of phytoplankton is always. lesser to the biomass of fishes which are dependent upon these phytoplankton.
3. The pyramid of biomass, therefore, in the sea is inverted.

Question 4.
The pyramid of energy is always upright.
Answer:

1. When the energy is moving from one trophic level to the next one, there is loss of some energy in the form of heat.
2. Therefore, the energy is always more on the lower trophic levels as compared to the higher trophic levels.
3. Due to this reason, the pyramid of energy is always upright and never inverted.

Question 5.
Only a fraction of sunlight is used for photosynthesis.
Answer:

1. When sunlight falls on the earth surface about 34% of this is reflected back.
2. About 10% is held by the ozone layer, water vapour and other atmospheric gases.
3. Out of the total solar energy, about 56% reaches to the earth’s atmosphere.
4. Only 0.02% of the sunlight is used for photosynthesis. This shows that only a fraction of sunlight is used for photosynthesis.

Question 6.
Usually crustose lichens form a pioneer species.
Answer:

1. The pioneer species is the one which invades a bare area.
2. When primary succession occurs it takes place on rocks.
3. The crustose lichens can secrete acids which can dissolve rocks. This starts weathering of rocks and soil formation.
4. Further, bryophytes and mosses can take hold of such soil. Therefore, as a pioneer on the barren rocks only crustose lichens can grow.

Write short notes

Question 1.
Zonation in wetland.
Answer:
There are four zones in wetland, viz. subtidal channels, mudflats, low marsh and high marsh. High marsh is more in terrestrial ecosystem while the subtidal channels are more of aquatic nature.

1. Subtidal channels : These are important habitat for fish during low tide. This region allows good drainage and flooding in mudflats.
2. Mudflats : This area is very rich in invertebrate life. Therefore, many wading birds are dependent for food in this area. Algal mat is also observed on mud flats.
3. Low marsh : This area is a good habitat for cordgrass, insects, herons and egrets and the clapper rails.
4. High marsh : This region supports pickleweed and patches of cordgrass. Sparrow and clapper rails are seen here.

Question 2.
Pond as a small ecosystem.
Answer:

1. For every ecosystem there are four important functional aspects viz. productivity, decomposition, nutrient cycling and energy flow.
2. In a small pond ecosystem, too, there are complex interactions.
3. Pond is a shallow water body in which all the above four basic processes of an ecosystem are observed.
4. The abiotic component is water along with dissolved inorganic and organic substances and rich soil deposit at the bottom of the pond.
5. The sunlight acts as a source of solar energy, the cycle of temperature, day-length and other climatic conditions regulate the rate of function of the entire pond.
6. Phytoplankton are the main producers along with algae and other aquatic plants.
7. Zooplankton, aquatic insects and fish are the consumers.
8. The decomposers are the fungi and bacteria present in the bottom soil.

Question 3.
Primary succession.
Answer:

1. Primary succession means the initial development of life on barren piece of land.
2. Primary succession occurs on newly cooled lava, rocks and newly created pond or reservoir. In a newly formed volcanic island, the life starts and takes up millions of years to develop the whole biomass.
3. In the successive serai stages, there is a change in the diversity of species of organisms, increase in the number of species and organisms as well as an increase in the total biomass.
4. The establishment of a new biotic community is generally very slow and takes millions of years.
5. Primary succession depends on climatic condition, soil characteristics and natural processes.

Question 4.
Secondary succession.
Answer:

1. Secondary succession begins in areas where previously natural biotic communities were present. But later were destroyed due to causes such as abandoned farm lands, burnt or cut forests, flooded lands, etc.
2. Secondary succession is faster than primary succession because already there is some soil or sediment present.
3. It occurs in the form of changed vegetation. But due to changed vegetation, there is also change in the resident animals that depend for food and shelter.
4. Thus, with secondary succession, the numbers and types of animals and decomposers also change.
5. During succession, natural or human induced disturbances (fire, deforestation, etc.), can convert a particular serai stage of succession to an earlier previous / preceding stage.
6. Sometimes, disturbances like this can create new conditions that encourage some species and eliminate other species.

Question 5.
Succession of Plants.
Answer:

1. Succession of plants is of two types based on the nature of the habitat. These are hydrarch or hydrosere which is based on water or very wetland areas and xerarch or xerosere based on very dry areas.
2. In wetter areas hydrarch succession begins and then successional series progress from hydric to the mesic conditions.
3. Xerarch succession starts in dry areas and the series progress from xeric to mesic conditions.
4. Both hydrarch and xerarch successions lead to mesic or medium water condition which is neither too xeric nor too hydric.

Question 6.
Pioneer species.
Answer:

1. Pioneer species are the ones that invade a bare area during primary succession.
2. Crustose lichens which are able to secrete acids to dissolve rock usually start as pioneer species. They bring about weathering of rocks and soil formation.
3. Later here bryophytes, mosses are settled as they can take hold in the small amount of soil. They are then followed by herbaceous plants, and after several more stages, ultimately a stable climax forest community is formed.

Question 7.
Succession in aquatic habitat.
Answer:

1. In aquatic habitats the pioneer species in primary succession are the small phytoplankton.
2. Phytoplankton are replaced by rooted- submerged plants (e.g. Hydrilla), rooted- floating angiosperms (e.g. Lotus) followed by free-floating plants (e.g.Pistia), then reed swamp (e.g. Typha), marsh-meadow (e.g. Cyperus), scrub (e.g. Alnus) and finally the trees (e.g. Quercus) in a very systematic and gradual way.
3. The climax again would be a forest. With passage of time, the water body is converted into land.

Short Answer Questions

Question 1.
Describe the functions of the ecosystem.
Answer:
Functions of an ecosystem:

1. Biological energy flow.
2. Productivity of the ecosystem.
3. Photosynthesis and respiration that take place in the ecosystem.
4. Nutrient cycles operating in the ecosystem.
5. Regulation of the environment by the organisms and the regulation of the organisms by the environment in turn.
6. Interactions of biotic and abiotic components of the ecosystem.

Question 2.
What are the types of ecosystems? Give their suitable examples.
Answer:
Ecosystems are of two types, viz. natural ecosystem and artificial ecosystem.
(1) Natural ecosystems : The ecosystems which operate under natural conditions without any much major interference by man are called natural ecosystems. E.g. terrestrial ecosystems such as grasslands, forests, deserts, etc. or aquatic ecosystems such as lakes, river, wetland, etc.

(2) Artificial ecosystems : The man engineered ecosystems which are maintained artificially by man by the addition of energy are called artificial ecosystems. These ecosystems are dependent upon manipulations from human beings. E.g. croplands, aquarium, aquaculture, etc. are the types of artificial ecosystems.

Question 3.
What are the different components of ecosystem?
Answer:
(1) There are two main components of the ecosystem, viz. abiotic components and biotic components.

(2) Abiotic components include all the inorganic substances such as B S, C, N, H, etc., their distribution and amount available in an ecosystem and the organic compounds such as proteins, carbohydrates, lipids, etc. along with the climate of that region.

(3) Biotic components include all the living organisms living in an ecosystem. The living organisms may be autotrophic or producers and converters or transducers. These are either green plants having photosynthetic abilities or chemosynthetic microorganisms. The heterotrophic organisms are called consumers.

They can either be macroconsumers or microconsumers.

1. Macroconsumers are herbivores, carnivores or omnivores as per the order in which they appear in the food chain.
2. Herbivores are primary consumers while the carnivores are secondary consumers.
3. Omnivores can be secondary or tertiary consumers.
4. Microconsumers are decomposing organisms and hence they are also called decomposers. They have saprophytic mode of nutrition. Bacteria, actinomycetes and fungi are some of the microconsumers.
5. Microconsumers decompose complex organic compounds from dead and living protoplasm and release the inorganic nutrients obtained from them, back to the environment.

Question 4.
What are the two spatial patterns in an ecosystem? Describe them briefly.
Answer:

1. There are two spatial patterns recognized in an ecosystem, viz. zonation and stratification.
2. Zonation is the spatial pattern which occurs horizontally along the ground. Along a horizontal gradient, the density and distribution of the species keep on varying.
3. Stratification is the spatial pattern which occurs vertically. This is determined by the height of organisms. Such stratification is seen in forest community where trees of different species grow to different heights.

Question 5.
Describe the concept of energy flow.
OR
In an ecosystem the flow of energy is unidirectional. Explain.
Answer:
(1) For considering the energy flow in an ecosystem the following aspects have to be taken into account:

• The efficiency of the producers in the absorption and conversion of the solar energy.
• The quantity of converted energy is used by the consumers.
• The total input of energy in the form of food and its efficiency of assimilation by the consumers.
• The amount of energy is lost through respiration, heat, excretion, etc.
• The ultimate gross net production.

(2) The energy captured by autotrophs never returns back to the sun. The energy obtained by the herbivores will never go back to autotrophs. Hence energy flow is always unidirectional.

(3) The energy flow through different trophic levels is progressive and hence previous trophic level cannot get this back.

(4) The amount of energy keeps on decreasing as it travels to further trophic level. This loss of energy is due to dissipation as heat formed during various metabolic activities of the organism. The energy loss is measured as respiration coupled with unutilized energy.

(5) If the food chain is shorter there is greater amount of available food energy. When the length of food chain increases, there is corresponding more loss of energy.

Question 6.
What could be the reason for the low productivity of ocean?
Answer:

1. The conditions of land and in ocean are not same. Primary productivity which is the biomass or dry weight produced by the plants per unit area during the photosynthesis is different for oceanic ecosystem.
2. In ocean, sunlight does not penetrate uniformly at all depths. It is thus the main limiting factor which decreases the rate- of photosynthesis.
3. Decrease in rate of photosynthesis decreases the growth of phytoplankton and then that of the zooplankton. Aquatic plants such as algae are also affected due to lack of sunlight.
4. Minerals and nutrients is also a limiting factor based on location of the oceans.
5. Therefore, there is less productivity of the oceans to about 55 billion tons as compared to productivity on land which is about 170 billion tonnes.

Question 7.
What could be the connecting points between the GFC and DFC?
Answer:

1. In grazing food chain, the energy is transferred from producers to consumers.
2. The autotrophs convert inorganic matter into organic compounds which is then transferred in food chain or food web.
3. In detritus food chain energy is obtained from organic matter or detritus generated in trophic levels of the grazing food chain.
4. Detritus such as dead bodies of animals or fallen leaves, which are then eaten by decomposers or detritivores.
5. These detritivores can also be consumed in grazing food chain by secondary consumers or predators.
6. When decomposers convert organic matter back into inorganic substances, the autotrophs use these minerals again.
7. In this way GFC and DFC are connected to each other in cyclic and mutual exchanges.

Question 8.
How will you classify man as carnivore (primary/ secondary) or omnivore? Why?
Answer:
Autotrophic producers or plants make the first trophic level. They synthesize their own food. Herbivores are primary consumers. They eat producers. Carnivores are secondary consumers. They eat primary consumers. The trophic level of man is omnivore as he eats both plants and animals. Non vegetarian person can be called carnivore while totally vegetarian person can be herbivore. However, the most appropriate placement of man is omnivorous.

Question 9.
How many trophic levels human beings function in a food chain?
Answer:
(1) All food chains and webs have at least two dr three trophic levels. Generally, there are a maximum of four tropic levels. Many consumers feed at more than one tropic level.

(2) Man is a primary consumer when he eat plants such as vegetables. Vegans and vegetarians who do not consume any animal product, fish or meat of any kind are primary consumers. They belong to the second trophic level.

(3) But some humans have different dietary choices. Many humans are omnivores, meaning they consume both plant and animal material. Thus, they may be on the third or even fourth tropic level.

(4) For example, if man consumes goat meat he is a part of the third tropic level.

(5) But when he eats bigger fish (and considering fish eats smellier fish) he becomes tertiary consumer on the fourth tropic level.

Question 10.
What is climax community? What leads to such a community?
Answer:

1. Climax community is the community which is near equilibrium with the environment.
2. The sequential changes in the structure and composition of the community, to adjust with the changing environment is called a climax community.

Question 11.
From algae to forest, explain in relation with the succession.
Answer:

1. When there is a succession from algae to forest, it depends upon the amount of water available.
2. The succession begins with small phyto-planktons followed by submerged and free floating and then rooted hydrophytes, sages, grasses and finally the trees.
3. Similarly, there is also a transformation from a pool of water to swamp then marsh and then mesic which means neither too dry nor too wet conditions.
4. Then small plants like mosses can inhabit followed by herbs, shrubs and then trees. Such succession ultimately leads to a stable climax forest community.

Question 12.
Explain the following terms with reference to ecological succession.
(1) Serai stages.
Answer:
The developmental stages of the ecological succession are known as serai stages.

(2) Pioneers.
Answer:
The organisms belonging to first serai stage in the ecological succession are known as pioneers.

(3) Hydrosere.
Answer:
Hydrosere or hydrarch succession is a type of ecological succession which is determined by the amount of water available during succession. Hydrosere occurs when there is abundant water available in the area where organisms reside.

(4) Xerosere.
Answer:
Xerosere or xerarch succession is a type of ecological succession which is determined by the amount of water available during succession. Xerosere occurs when there is very little water available in the area where organisms reside. Such succession is observed in desert regions.

Question 13.
Why is zonation more pronounced at the edges of habitat?
Answer:

1. Edge of the habitat is an abruptly changing region. E.g. the shore or littoral zone is the edge between aquatic and terrestrial habitats.
2. At such places there is variety of environmental factors, such as temperature, wind exposure, light intensity, wave action, and salinity, etc.
3. These abiotic factors are never constant in such areas and they show tremendous variation.
4. Therefore, the intertidal communities show differences in regions that are occupied by them. In this way zonation is more pronounced in such areas when edges of habitat are present.

Question 14.
What is the maximum number of trophic levels in a food chain?
Answer:

1. There are maximum four trophic levels in an ecosystem.
2. Rarely five levels are seen where it is occupied by apex consumer.
3. But as the trophic levels are moving from producers to consumers, lesser and lesser energy remains.
4. Through heat loss, lot of amount of energy is dissipated, therefore no ecosystem can sustain fifth trophic level.

Question 15.
How would you explain inverted pyramid of biomass in oceanic ecosystem?
Answer:

1. Pyramid of biomass is inverted in an oceanic ecosystem because in oceans the biomass of fishes is much more than the biomass of phytoplanktons.
2. The biomass of phytoplanktons which are the producers in the ocean is smaller than that of zooplanktons.
3. Zooplankton are primary consumers in oceans. The biomass of zooplanktons is smaller than that of secondary consumers which are fish.
4. This results in the inverted pyramid of biomass is an oceanic ecosystem.
5. Also the life span of phytoplankton is very small, which are consumed almost as rapidly as they are formed.
6. The phytoplankton have high reproductive potential by which they reproduce rapidly. All these facts make the pyramid inverted in case of oceanic ecosystem.

Question 16.
What is secondary productivity? What is annual net primary productivity of biosphere?
Answer:

1. Secondary productivity is defined as the rate of formation of new organic matter by consumers.
2. It is the rate of assimilation of food energy by the consumers.
3. This amount of energy available to consumer for transfer to the next trophic level.
4. The annual net primary productivity of the whole biosphere is approximately 170 billion tonnes (dry weight) of organic matter. Of 170 billion tonnes, the oceans create about 55 billion tonnes, and rest is by land ecosystems.

Complete the given chart

Trophic levels	Type of organisms	Status	Example
…………….….	……………………..	Secondary consumer	……………………..
……………..…	Herbivore, Heterotrophs	……………………..	…………………….
……………..…	Autotrophs	…………………….	Phytoplankton, grass, trees

Answer:

Trophic levels	Type of organisms	Status	Example
Third trophic level	Carnivore, Heterotrophs	Secondary consumer	Lion, Frog
Second trophic level	Herbivore, Heterotrophs	Primary consumer	Cattle, Sheep, Deer
First trophic level	Autotrophs	Producer	Phytoplankton, grass, trees

Diagram based questions

Question 1.
Sketch and label Carbon cycle
Answer:

Question 2.
Sketch and label phosphorus cycle
Answer:

Question 3.
Sketch and label decomposition cycle
Answer:

Question 4.
Sketch and label simple grazing food chain
Answer:

Question 5.
Sketch and label ecological pyramids of energy and pyramid of biomass
Answer:

Question 6.
Observe pyramids of numbers and answer the questions below

Questions:
1. In the above pyramid of numbers how many primary consumers are supporting secondary consumers? What will happen if the numbers are reversed?
2. When does pyramid of numbers get inverted in case of a single tree ecosystem ?
3. What will happen, if in the above example, we substitute larger bird of prey feeding on small insect eating birds?
Answer:
(1) 500,000 primary consumers are supporting 5000 secondary consumers. If the numbers are reversed, i.e. if primary consumers are lesser than secondary consumers, then the secondary consumers will have fierce competition and will lead to decline in their number too. The pyramid of numbers will get inverted in such case.

(2) If we plot the number of insects on a single tree, smaller birds feeding on insects, and parasites on those birds, we get an inverted pyramid.

(3) If large birds are feeding on smaller insect eating birds, their population will decrease. This will result into increased number of insects as there will be no check on the insect population due to loss of smaller predator birds. If larger birds keep on feeding constantly and unchecked on smaller birds, the smaller ones will eventually become lesser and lesser in their population and this in turn will starve the larger birds too, leading in decrease of their population too.

Question 7.
Observe the given food web and answer the questions

Questions
1. Name the producers in the above web.
2. Name the primary consumers in the above web.
3. Which is an omnivore animal in the above web? Why?
4. Why frog is occupying an important ecological position in this web?
5. From the given food web diagram, give the trophic levels where the eagle is present.
Answer:
1. Corn crop, flowering plant, lavender plant, mango tree
2. Grasshopper, butterfly, fruit fly
3. Rat, because it feeds on grains of corn as well as on insects.

4. Frog feeds on insects and keeps the insect population under check. It also falls prey to snakes and birds of prey like eagle. Thus, supports their population by providing food.

5. (1) Eagle is apex consumer in the following cases : Lavender/Producer → Butterfly/ Primary consumer → Dragon fly /Secondary consumer → Thrush/Tertiary consumer → Eagle/Apex consumer
Or
Corn/Producer → grasshopper/Primary consumer → frog/Secondary consumer → python/Tertiary consumer → eagle/Apex consumer.
(2) Eagle is tertiary consumer in the following case : Rat/Primary consumer → python/ Secondary consumer → eagle/Tertiary consumer
(3) Secondary consumer in the following case : Rat/Primary consumer → eagle/ Secondary consumer

Question 8.
Observe the given figure and answer the questions:

Questions:
1. Which trophic level has maximum energy?
2. Why carnivores have least energy shown in this diagram?
Answer:
1. Producers which form first trophic level has maximum energy.

2. When energy flows from one trophic level to the next, some amount is lost as heat. As it can be seen that producers had 1000 joules of energy, out of which 900 joules are lost when energy flowed from producers to primary consumers. Further, it was lost again by 90 joules while herbivore to carnivore energy flow was taking place. Therefore, the carnivore gets least energy as it is the higher trophic level.

Question 9.
Identify A, B, C, D and E given in sketch of carbon cycle

Answer:
A : Combustion
B : Respiration by plants
C : Respiration by animals
D : Photosynthesis
E : Decomposition.

Question 10.
Observe the diagram and answer the questions

1. Fill in the empty boxes with proper words in the above figure.
2. Which cycle is it depicting?
3. What kind of cycle is it?
Answer:
1. A : geological uplifting, B : Weathering of phosphate from rocks, C : Phophaste in solution. D : Detritus settling at bottom, E : Sedimentation forming New rocks, F : Phosphate in soil, G : Leaching
2. Phosphorus cycle
3. Sedimentary cycle

Long answer questions

Question 1.
Describe the pathway followed by phosphorus in an ecosystem.
Answer:

(1) The phosphorus cycle is the simplest of all the nutrient cycles, which is a type of sedimentary cycle. It constitutes cyclic movement of phosphorus through hydrosphere, lithosphere and biosphere.

(2) Since phosphorus is a heavy molecule it never goes into the atmosphere. Phosphorus remains in the bodies of organisms, dissolved in water or in the form of rock. The natural reservoir of phosphorus is rock in which phosphorus is present in the form of phophates.

(3) Upon weathering of the rock due to action of mildly acidic water, the phosphates in the rock go into the solution.

(4) Plants take up phosphorus in the form of phosphate. The roots of plants can absorb phosphate ions from the soil. Animals obtain phosphorus through food which they consume. Thus autotrophs supply phosphorus to heterotrophs. Phosphorus is a major biological constituent of all the living organisms.
It is found in biological membranes, nucleic acids e.g. DNA, RNA, cellular energy transfer systems such as ATE Phosphorus is thus an essential element.

(5) The requirement of phosphorus in animals is much more as it is the component of bones, hooves, teeth, shells.

(6) The waste products formed through defecation and the dead organisms are decomposed by phosphate-solubilizing bacteria releasing phosphorus. This in turn is used up by other growing plants. Phosphorus is always in short supply and hence acts as limiting factor for the plant growth.

Sudden influx of phosphorus in the form of agricultural runoff or industrial effluents rich in phosphate content, leads to eutrophication in water bodies. Eutrophication is due to overgrowth of algae at the instance of high phosphorus dissolved in water. The overgrowth of algae kills or harms the aquatic life.

Question 2.
What are the most important ecological services whose value cannot be determined?
Answer:
(1) The main ecological services are fixation of atmospheric CO₂ and release of O₂ are the most important services provided by an ecosystem.

(2) Photosynthetic activity of photoautotrophs helps in carbon sequestration in the form of CO₂ from the atmosphere. At the same time it releases O₂ as a by-product.

(3) O₂ is needed for respiration by all aerobic organisms. Oxygen also purifies air.

(4) For human consumption, crops and fruits are continuously required. These can be produced only after pollination of plants. Wind, water or other biotic agent such as insects therefore play an important role as ecosystem service. Without pollination there would be no crops and no fruits.

Question 3.
What are the four categories of ecosystem services as given by Millennium Ecosystem Assessment Report in 2005? What are the services included in each?
Answer:
Ecosystem services are of following four categories:
(1) Supporting services : Support to the life on earth such as nutrient cycling, primary production, soil formation, habitat provision, pollination and overall maintenance of balance of ecosystem.

(2) Provisioning services : Provides necessities such as food in the form of crops, fruits and seafood, raw materials such as timber, skins, fuel wood, genetic resources in the form of seeds, crop improvement genes, and health care, other resources such as water, medicinal resources in the form of test and assay organisms and ornamental resources such as furs, feathers, ivory, orchids, butterflies, etc.

(3) Regulating services : Regulation of processes on the earth such as carbon sequestration, prey-predation regulation, waste decomposition and detoxification, purification of water and air and pest control.

(4) Cultural services : Under this category, humans get services from nature in the form of cultural, spiritual and historical, recreational experiences, opportunity to learn science and indulge in education, and pets and animal therapy.

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 9 Morphology of Flowering Plants Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 9 Morphology of Flowering Plants

Question 1.
Explain how angiosperms are classified into different types based on habitat.
Answer:
Angiosperms can be classified into following types based on habitat:

1. Hydrophytes – Growing in aquatic habitat e.g. Hydrilla
2. Xerophytes – Growing in regions with scanty or no rainfall like desert e.g. Opimtia
3. Psammophytes – Growing in sandy soil e.g. Elymus
4. Lithophytes – Growing on rock e.g. Couchidium, Cladopus, Dalzellia, Paphiopedilum orchids, rock felt fem.
5. Halophytes – Growing in saline soil e.g. Mangrove plants like Rhizophora

Question 2.
Draw neat and labelled diagram of a typical angiospermic plant. Classify its vegetative and reproductive structures.
Answer:

Vegetative structures in angiospermic plant	Reproductive structures in angiospermic plant
Root, Stem, Leaf	Flowers, Fruits, Seeds

Question 3.
Label the various regions of a typical root in the given figure and explain them in detail.
Answer:

A typical root possesses the following regions:
1. Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root.

2. Meristematic region or region of cell division:
a. The apex of the root is a growing point about 1 mm in length protected by root cap. This region is called as region of cell division or meristematic region.
b. The structure is developed by compactly arranged thin walled actively dividing meristematic cells.
c. These cells bring about longitudinal growth of root.

3. Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

4. Region of root hair or region of absorption:
a. A region of root hair / absorption/piliferous zone is made up of numerous hair like outgrowths.
b. The epiblema or piliferous layer produces tubular elongated unicellular structures known as root hair.
c. They are in close contact with soil particles and increase surface area for absorption of water.
d. Root hair are short lived or ephimeral and are replaced after every 10 to 15 days.

5. Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question 4.
What are the primary functions of root?
Answer:
Primary functions of root are, fixation or anchorage of plant body in the soil, absorption of water and minerals from soil and conduction of absorbed materials up to the stem base, etc.

Question 5.
Which type of root system is found in plants like maize, wheat and sugarcane? Explain in detail.
Answer:

1. Adventitious root system is found in plants like maize, wheat and sugarcane.
2. Adventitious root develops from any part other than radicle.
3. Such roots may develop from the base of the stem, nodes or from leaves.
4. In monocots, radicle is short lived.
5. A thick cluster of equal sized roots arise from the base of a stem. It is also known as fibrous root system as they look like fibre. The growth of roots is superficial.
6. Adventitious roots in some plants are used for vegetative propagation. E.g. Euphorbia, Carapichea ipecacuanha (Ipecac) etc.

Question 6.
What are metamorphosed roots?
Answer:
When roots have to perform some special type of function in addition to or instead of their normal function they develop some structural changes. Such roots are called as metamorphosed roots.

Question 7.
Complete the given chart and explain the modification of tap root for storage of food.
Answer:

1. Modification of tap root for storage of food:
a. WTien tap root stores food it becomes swollen, fleshy and also develops definite shape.
b. Main or primary root is the main storage organ but sometimes hypocotyl part of embryo axis also joins the main root. Secondary roots remain thin.
c. On the basis of shape, swollen tap roots are classified as Fusiform, Conical and Napiform.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)
3. Conical root:
The conical root is broad at its morphological base and narrows down towards its apex. e.g. Carrot (Daucus car ota)
4. Napiform root:
In napiform root, base of root is highly swollen, almost spherical in shape and abruptly narrows down towards its apex. e.g. Beet (Beta vulgaris)

Question 8.
Identify the type of swollen tap root in the figures given below.

Answer:
Figure ‘a’: Conical root;
Figure ‘b’: Fusiform root;
Figure ‘c’: Napiform root

Question 9.
Answer the following questions:
1. Identify the label ‘X’ in the given figure of respiratory roots. Give its function.

2. Give any tw o examples of plants in which respiratory roots are present.
Answer:
1. X: Lenticels
In respiratory roots or pneumatophores, gaseous exchange occurs through lenticels.
2. Examples of plants in which respiratory roots are present:
Rhizophora, Avicennia, Sormeratia, Heritiera fames (sundri), etc.

Question 10.
Identify the types of modified adventitious roots in the figures given below and explain in detail.

Answer:
1. Figure ‘a’ indicates simple tuberous root.
a. Simple tuberous roots become swollen and do not show definite shape.
b. They are produced singly.
c. The roots arise from nodes over the stem and penetrate into the soil, e.g. sweet potato or shakarkand (Ipomoea batatas).

2. Figure ‘b’ indicates fasciculated tuberous roots.
a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

3. Figure ‘c’ indicates Moniliform roots.
a. Some adventitious roots get swollen at regular intervals.
b. These gives them the appearance of beads of a necklace. Such roots are called as Moniliform roots, e.g. Spinacia oleracea (Indian Spinach).

4. Figure ‘d’ indicates Nodulose roots.
The cluster of long slender roots become enlarged at the tips forming nodules is known as nodulose roots, e.g. Arrow root (Maranta), Amhaldi or mango ginger (Curcuma amada).

Question 11.
Explain various types of adventitious roots which are modified for mechanical support.
Answer:
1. Prop roots / Columnar roots:
a. These roots arise from horizontal branches of tree like Banyan tree (Ficus benghalensis) and grow vertically downwards till they penetrate the soil.
b. These prop roots show secondary growth, become thick, act like pillars to provide mechanical support to the heavy branches.

2. Stilt roots:
a. These roots normally arise from a few lower nodes of a weak stem in some monocots, shrubs and small trees.
b. They show obliquely downward growth penetrating soil and provide mechanical support to the plant.
c. In the members of family Poaceae, the plants like Maize, Jowar, Sugarcane etc. produce stilt root in whorl around the node.
d. These roots provide additional support to the plant body.
e. In Screw pine or Pandanus (Kewada), stilt roots arise only from the lower surface of obliquely growing stem for additional support. These roots show multiple root caps.

3. Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

4. Clinging roots:
a. These tiny roots develop along intemodes, show disc at tips, which exude sticky substance.
b. This substance enables plant to get attached with walls of buildings.
c. They do not damage substratum, e.g. English Ivy (Hedera helix).

5. Plank roots/Buttress:
a. These roots often develop at the base of large trees and form plank like extensions around stem.
b. These roots provide additional support, e.g. Silk cotton, Peepal, etc.

6. Buoyant roots:
Roots developed at the nodes of aquatic herbs like (Jussiaea repens), become highly inflated and spongy providing buoyancy and helping the plant to float.

Question 12.
What are sucking roots? Explain with the help of examples.
Answer:
Sucking roots or Haustoria:
1. These are the specialised microscopic sucking roots developed by parasitic plants to absorb nourishment from the host.
2. Viscum album is a partial parasite. It develops haustoria which penetrate into xylem of host plant for absorption of food.
3. In Cuscuta reflexa or Dodder (Amarvel) haustoria penetrates vascular strand and suck food from phloem, water and minerals from xylem. Cuscuta is leafless plant with yellow stem. It is a total parasite.

Question 13.
Enlist the important characteristics of stem.
Answer:
Characteristics of stem:

1. Stem is the ascending part of the plant body which develops from plumule and reproductive units.
2. It is usually positively phototropic, negatively geotropic and negatively hydrotropic.
3. It shows different types of buds (axillary, apical, accessory, etc.).
4. It is differentiated into nodes and intemodes.
5. At nodes it produces dissimilar organs such as leaves and flowers and similar organs such as branches.
6. Young stem is green and capable of photosynthesis.

Question 14.
Sketch and label a typical stem structure and write its primary functions.
Answer:

The primary functions of the stem are to produce and support branches, leaves, flowers and fruits; conduction of water and minerals and transportation of food to plant parts.

Question 15.
What is underground stem? When does it produce aerial shoots?
Answer:
In some herbaceous plants the stem which develops below soil surface is called underground stem. The underground stem remains dormant during unfavourable condition and on the advent of favourable condition produces aerial shoots.

Question 16.
Draw neat and labelled diagram of rhizome of ginger.
Answer:

Question 17.
Potato which we eat is an underground part of a plant, however it can not be considered as root. Justify the given statement.
Answer:

1. Potato is a stem tuber.
2. It is an underground stem, modified for storage of food material.
3. Special underground branches of stem at their tips becomes swollen due to storage of food which is mostly starch.
4. Stem tuber shows distinct nodes, but not intemodes hence it is classified as stem.
5. At nodal part, it shows scale leaves with axillary buds, which are commonly called as ‘eyes’.
6. Under favourable conditions, ‘eyes’ can produce aerial shoots.
7. Potato tuber can be propagated vegetatively. [Note: In stem tuber, internodes are present but they are not very distinct.]

Question 18.
What are tunicated and compound tunicated bulbs?
Answer:
1. Tunicated bulb:
When fleshy scale leaves are arranged on stem in concentric manner, bulb is called as tunicated bulb or layered bulb. E.g. Onion
2. Compound tunicated bulb:
When fleshy scale leaves arranged on stem, partially overlap each other by their margins only, such bulb is called compound tunicated or scaly bulb. e.g. Garlic

Question 19.
Which type of modified stem is present in Colocasia and Amorphophallus? Explain with the help of neat and labelled diagram.
Answer:

1. In Colocasia and Amorphophallus corm is present, which is an underground stem modified for storage of food.
2. Corm is swollen underground spherical or subspherical vertically growing stem.
3. It is condensed structure with circular or ring like nodes.
4. It shows presence of axillary buds and scales.
5. Adventitious buds are produced which help in vegetative propagation.
6. Adventitious roots are produced at lower part of the stem.

Question 20.
1. What are sub aerial stems?
2. Explain the different types of sub aerial stems. Give atleast one example of each.
Answer:
1. Subaerial stems:
a. These are generally weak or straggling stems growing over the ground and need support for perpetuation.
b. Sometimes these stems are found to grow beneath the soil surface also. Thus, they show contact with both air and soil.
c. Subaerial stems are meant for perennation and vegetative propagation.
d. Scale leaves and axillary buds are present over stem surface. Axillary buds develop into aerial shoots.

2. Types of subaerial stems:
a. Trailer:
1. The shoot spreads over the ground without striking adventitious roots.
The branches are either flat i.e. procumbent or partly vertical i.e. decumbent.
e. g. Euphorbia, Tridax etc. [Any one example]

b. Runner:
1. They are special narrow, prostrate or horizontal green branches which develop at the base of erect shoots known as crown.
2. Runners spread in all directions to produce new crowns with bunch of adventitious roots.
3. Presence of nodes with scale leaves and axillary buds is observed.
e.g. Cynodon (Lawn grass) Centella (Hydrocotyl / Brahmi), Oxalis etc. [Any one example]

c. Stolons:
1. The slender lateral branch arising from the base of main axis is known as stolon.
2. In some plants it is above ground (wild strawberry).
3. Primarily stolon shows upward growth in the form of ordinary branch, but when it bends and touches the ground terminal bud grows into new shoot and develops adventitious roots.
e.g. Wild Strawberry, Jasmine, Mentha, etc. [Any one example]

d. Sucker:
1. It is non-green, runner like branch of stem.
2. It grows horizontally below soil initially and then comes above the soil surface obliquely to produce a new plant.
3. Sucker can be termed as underground runner.
e. g. Chrysanthemum, Banana etc. [Any one example]

e. Offset:
1. These are one intemode long runners in rosette plants at ground or water level.
2. Offset helps in vegetative propagation.
e.g. Water hyacinth or Jal kumbhi (Eichhornia) and Pistia. [Any one example]

Question 21.
Observe the given figures and identify P, Q, R and S representing the different types of sub aerial shoot.
Answer:
‘P’: Trailer ‘Q’: Runner ‘R’: Stolon ‘S’: Offset

Question 22.
Draw neat and labelled diagram of Eichhornia showing offset.
Answer:

Question 23.
What are metamorphosed stems?
Answer:
Stem or its vegetative part develops various modifications to carry out specialized functions. Such modified stems are called as metamorphosed stems.

Question 24.
Describe the aerial modifications of stem.
Answer:
Different aerial modifications shown by stem are as follows:
1. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.
d. Apical bud in Vitis quadrangularis gets modified into tendril. The further growth is carried out by axillary bud.
e. In Passiflora axillary bud gets modified in tendril.
f. Extra axillary bud is the one which grows outside the axil. This bud in cucurbita gets modified into tendril.
g. Normally floral buds are destined to produce flowers. But in plants like Antigonon they produce tendrils.

2. Thorn:
a. It is modification of apical or axillary bud.
b. Thom is hard pointed and mostly straight structure (except Bougainvillea where it is curved and useful for climbing).
c. It provides protection against browsing animals and also helps in reducing transpiration.
d. Apical bud develops into thorn in Carrisa whereas axillary bud develops into thorn in Duranta, Citrus, Bougainvillea, etc.

3. Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

4. Cladodes:
a. The branches of limited growth i.e. one intemode long and performing photosynthetic function are called as cladodes.
b. True leaves are reduced to spine or scales to reduce rate of transpiration, e.g. Asparagus.

5. Cladophylls:
These are leaf like structures bore in the axil of scale leaf. It has floral bud and scale leaf in the middle i.e. upper half is leaf and lower half is stem. e.g. Ruscus.

6. Bulbils:
a. In plants like Dioscorea, etc. axillary bud becomes fleshy and rounded due to storage of food called as bulbil.
b. When it falls off it produces new plant and help in vegetative propagation.

Question 25.
Observe the given figures oi Asparagus, Vitis quadrangularis and Passiflora. Which of the following is labelled incorrectly?
Answer:
Figure ‘a’ is incorrectly labelled.
It represents Cladode oi Asparagus.

Question 26.
How does thorn in Carissa differ from that of Duranta?
Answer:
In Carrisa, apical bud develops into thorn, whereas in Duranta, axillary bud develops into thorn.

Question 27.
Enlist the general characteristics of a leaf.
Answer:
General characteristics of a leaf:

1. Leaves are the most important appendages as they carry out photosynthesis and also help to remove excess amount of water from plant body through transpiration.
2. Leaves are exogenous in origin and develops from leaf primordium.
3. Leaf is dorsiventrally flattened lateral appendage of stem, produced at nodal region.
4. Leaf is thin, expanded and green due to presence of photosynthetic pigments, i.e. chlorophyll.
5. Axil of leaf shows presence of axillary bud.
6. Leaf shows limited growth, does not show apical bud or a growing point.

Question 28.
Give an account of various parts of a typical dicot leaf.
Answer:
A typical dicot leaf shows presence of three main parts Leaf base or Hypopodium, Petiole or Mesopodium and Leaf lamina/blade or Epipodium.
1. Leaf base or Hypopodium:
a. The point by which leaf remains attached to the stem is known as leaf base.
b. The nature of leaf base varies in different plants. It may be pulvinus (swollen), sheathing or ligulate, etc.

Question 29.
How simple leaf differs from a compound leaf?
Answer:
The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.

Question 30.
Identify the type of pinnately compound leaves in the figures given below and give one example of each.
Answer:

Figure ‘a’: Paripinnately compound leaves (e.g. Cassia)
Figure ‘b’: Imparipinnately compound leaves (e.g. Rosa)
Figure ‘c’: Bipinnately compound leaves (e.g. Caesalpinia)
Figure ‘d’: Tripinnately compound leaves (e.g. Moringa)
Figure ‘e’: Decompound leaves (e.g. Coriandrum)

Question 31.
Explain how leaves modify to perform different functions other than photosynthesis and gaseous exchange.
Answer:
Leaves show different types of modification as follows:
1. Leaf spines:
Sometimes entire leaf is modified into spines (Opuntia) or margin of leaf becomes spiny (Agave) or stipule modifies into spine (Acacia, Zizyphus) to check the rate of transpiration and to protect plant from grazing.

2. Leaf tendril:
In some weak stems, leaf, leaflet or other part modifies to produce thin, green, wiry, coiled structure called as leaf tendril. It helps in climbing and provides additional support.

3. Leaf hooks:
In plants like Bignonia unguis-cati (Cat’s nail) the terminal three leaflet get modified into three! stiff curve and pointed hooks used to cling over the bark of tree.

4. Phyllode:
When petiole of leaf becomes flat, green and leaf like it is called as phyllode. In Acacia auriculoformis the normal leaf is bipinnately compound and falls off soon. The petiole modifies itself into phyllode. It is a xerophytic adaptation.

Question 32.
Identify different types of tendril in the figures given below and give one example of each.

Answer:
Figure ‘a’: Whole leaf tendril (e.g. Lathyrus)
Figure ‘b’: Leaflet tendril (e.g. Pisum sativum)
Figure ‘c’: Leaf tip tendril (e.g. Gloriosa)
Figure ‘d’: Stipular tendril (e.g. Smilax)

Question 33.
Define phyllotaxy. Why do leaves show phyllotaxy?
Answer:
Phyllotaxy:
1. Arrangement of leaves on the stem and branches in a specific manner is known as phyllotaxy.
2. It enables leaf to get sufficient light for photosynthesis.

Question 34.
Which type of phyllotaxy is show n by leaves of Mango, Nerium and Jamun?
Answer:

1. Phyllotaxy shown by leaves of Mango is alternate phyllotaxy. In this type, single leaf from each node.
2. Phyllotaxy shown by leaves of Nerium is whorled phyllotaxy. In this type, many leaves arise from each node
   and form a whorl.
3. Phyllotaxy shown by leaves of Jamun is opposite superposed phyllotaxy. In this type, a pair of opposite leaves are arranged one above the other in the same plane.

Question 35.
What are pinnately compound and palmately compound leaves? Give any two examples of each.
Answer:
1. Pinnately compound: Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf. e.g. Cassia, Rose, Caesalpinia, Moringa, Coriandrum, etc. [Any two examples]
2. Palmately compound: In this all the leaflets are attached at the tip of petiole.
e.g. Citrus, Zorina, Oxalis, Marsilea, Bombax [Any two examples] [Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 36.
Define inflorescence.
Answer:
A specialised axis or branch over which flowers are produced or borne in definite manner is known inflorescence.

Question 37.
Write significance of inflorescence.
Answer:
Significance of inflorescence:

1. Inflorescence makes a flower more conspicuous to attract the insects and birds for pollination.
2. It provides more chances for cross pollination.
3. An insect can pollinate many flowers in inflorescence in a single visit.
4. In an inflorescence, flowers open successively and not simultaneously. This improves chances of pollination as flowering period is longer.

Question 38.
Define flower.
Answer:
Flower is highly modified and condensed shoot meant for sexual reproduction.

Question 39.
Complete the table by giving the meaning of following terminologies related to flower.
Answer:

1. Complete flower: Presence of all four floral whorls	Incomplete flower: Absence of any one of the floral whorls.
2. Pedicellate flower: Flower with pedicel	Sessile flower: Flower without pedicel
3. Bracteate flower: Flower with bract at the base of pedicel	Ebracteate flower: Flower without bract
4. Perfect flower: Both androecium and gynoecium are present, also called as dicliny or bisexual flower.	Imperfect flower: Any one reproductive whorl is present also called as monocliny or unisexual flower.
5. Actinomorphic flower: The flower can be cut in any plane passing through the centre in order to obtain two identical halves. Flowers show radial symmetry, e.g. Sunflower	Zygomorphic flower: The flower can be cut only along one plane passing through the centre in order to obtain two identical halves. Flowers show bilateral symmetry e.g. Sweet Pea flower.
6. Unisexual flower: It can be either staminate (male)/ pistillate (female) flower	Neuter flower: When both reproductive whorls are absent, it is said to be neuter flower
7. Monoecious plant: Male and female reproductive flowers are borne on same plant, e.g. Maize	Dioecious plant: Only one type of unisexual flowers are present on plant, e.g. Ray floret of sunflower

Question 40.
Mango is called as polygamous plant. Why?
Answer:
Mango produces all types of flowers like staminate, bisexual and neuter, hence it is called as polygamous plant.

Question 41.
What is insertion of floral whorls?
Answer:
The position and arrangement of rest of the floral whorls (calyx, corolla, androecium) with respect to gynoecium on the thalamus is known as insertion of floral whorls. .

Question 42.
With help of neat and labelled diagrams explain the classification of flowers based on the position of ovary on the thalamus.
Answer:
Classification of flowers based on the position of ovary on the thalamus.

1. Hypogynous flower:
When the convex or conical thalamus is present in flower, ovary occupies the highest position while other floral parts are below ovary. Ovary is said to be superior and flower is called as hypogynous flower, e.g. Brinjal, Mustard, China rose etc. It is denoted as G in floral formula.

2. Perigynous flower:
When cup shaped or saucer shaped thalamus is present in a flower, ovary and other floral parts occupy about same position. Such an ovary is said to be semi- superior or semi-inferior. All floral whorls are at the rim of thalamus. Flower is called as perigynous. e.g. Rose, etc. It is denoted as G- in floral formula.

3. Epigynous flower:
When thalamus completely encloses ovary and may show fusion with wall; the other floral parts occupy superior position and ovary becomes inferior. Such flower is called as epigynous flower, e.g. Ray florets of Sunflower, Guava, Cucumber etc. It is denoted as G in floral formula.

Question 43.
What is thalamus?
Answer:
Thalamus:
1. The upper, swollen, condensed, knob-like part of the pedicel is called thalamus. It is also called receptacle or torus.
2. In a typical flower, the thalamus consists of four compactly arranged nodes and three highly condensed intemodes.
3. From each node of thalamus, whorl of modified leaves is produced.

Question 44.
With the help of a diagram explain the floral parts of a typical flower.
Answer:

Floral parts of a typical flower:
1. Calyx (K):
a. It is outermost floral whorl and individual members are known as
sepals.
b. Sepals are usually green in colour and perform photosynthesis.
c. If all the sepals are united, the condition is gamosepalous and if they are free, the condition is called as polysepalous.
d. Gamosepalous calyx is found in China rose and polysepalous calyx is found in Brassica.
e. The main function of sepals is to protect inner floral parts in bud condition.
f. Sometimes sepals become brightly coloured (petaloid sepals) and attract insects for pollination,
e.g. Mussaenda etc.
g. Sepals modify into hairy structures called as pappus. Such calyx helps in dispersal of fruit, e.g. Tridex.

2. Corolla (C):
a. It is second floral whorl from outer side and variously coloured.
b. The individual member is called as petal.
c. Petals may be sweet to taste, possess scent, odour, aroma or fragrance etc.
d. The condition in which petals are free is said to be polypetalous (e.g. Rose) and if they are fused it is called as gamopetalous (e.g. Datura).
e. The main function of corolla is to attract different agencies for pollination.

3. Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

4. Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.
2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.
3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.

5. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.
1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 45.
Explain the term Epicalyx.
Answer:
Epicalyx:

1. Epicalyx is an additional whorl of sepal like structures formed by bracteole which occurs on the outside of calyx.
2. These are 5-8 in number.
3. It is a characteristic feature of family Malvaceae.
4. They are protective in function, e.g. Ladies finger

Question 46.
Match the columns.

Column I (Tvpe of calyx)	Column II (Nature of sepals)	Column III (Example)
1. Caducous	(a) Sepals remain even after fruit formation	1. Brinjal, Pea
2. Deciduous	(b) Sepals fall off as soon as the flower bud opens	2. Lotus, Mustard
3. Persistent	(c) Sepals survive till (withering of petals) fruit formation	3. Argemone (Poppy)

Answer:

Column I (Type of calyx)	Column II (Nature of sepals)	Column III (Example)
1. Caducous	(b) Sepals fall off as soon as the flower bud open	3. Argemone (Poppy)
2. Deciduous	(c) Sepals survive till (withering of petals) fruit formation	2. Lotus, Mustard
3. Persistent	(a) Sepals remain even after fruit formation	1. Brinjal, Pea

Question 47.
Define aestivation and explain its different types.
Answer:
Aestivation:
1. The mode of arrangement of sepals, petals or tepals in a flower with respect to the members of same whorl is known as aestivation.
2. Different types of aestivation are as follows:
a. Valvate: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g. Calyx of Datura, Calotropis.
b. Twisted: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g. Corolla of China rose, Cotton etc.
c. Imbricate: One of the sepals or petals is internal and is overlapped at both the margins. One is external i.e. both of its margin overlap adjacent member. Rest of the sepals / petals have one inner or overlapped margin and outer or overlapping margin, e.g. Cassia, Bauhinia, etc.
d. Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

Question 48.
Answer the following:
1. Define the term Adelphy.
2. Give three types of stamens based on adelphy. Draw a diagram of each type.
Answer:
Adelphy:
1. When stamens are united by filaments and anthers are free, the condition is called adelphy.
2. Three types of stamens based on adelphy:
Monadelphous stamens, Diadelphous stamens, Polyadelphous stamens.

Question 49.
Define the following terms and give one example of each:

1. Epipetalous stamens
2. Epiphyllous stamens
3. Syngenesious stamens
4. Synandrous stamens

Answer:

1. Epipetalous stamens: When the stamens are united to petals they are described as epipetalous stamen. E.g. Datura
2. Epiphyllous stamens: When the stamens are united to tepals they are described as epiphyllous stamens. E.g. Lily
3. Syngenesious stamens: When anthers are united and filaments are free, such stamens are called as syngenesious stamens. E.g. Sunflower
4. Synandrous stamens: When both anthers and filaments are fused, such stamens are called as synandrous stamens. E.g. Cucurbita

Question 50.
Define placentation. Explain its types.
Answer:
1. Placentation: The mode of arrangement of ovules on the placenta within the ovary is called placentation.
2. Types of placentation:
a. Marginal: Ovules are placed on the fused margins of unilocular ovary, e.g. Pea, Bean etc.
b. Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.
c. Parietal: Ovules are placed on the inner wall of unilocular ovary of multicarpellary syncarpus ovary,
e. g. Papaya, Cucumber, etc.
d. Basal: Single ovule is present at the base of unilocular ovary, e.g. Sunflower, Rice, Wheat.
e. Free central: Ovules are borne on central axis which is not attached to ovary wall, e.g.Argemone, Dianthus.

Question 51.
What are parthenocarpic fruits? Give any two examples.
Answer:
Fruits which are produced from ovary without fertilization are called as parthenocarpic fruits, e.g. Cultivated Banana and Grapes.

Question 52.
How true fruit differs from false fruit or pseudofruit? Give one example of each.
Answer:
The fruit which develops only from ovary is called as true fruit or eucarp. e.g. Mango. The fruit which develops from other than ovary floral part is called as false fruit or pseudocarp

Question 53.
Identify the parts of mango and apple fruit in the given figures.
Answer:
1. parts of mango fruit:
X: Epicarp
Y: Mesocarp
Z: Endocarp

2. Part of apple fruit:
P: Thalamus
Q: Seed
R: Endocarp
S: Mesocarp

Question 54.
Identify terms a,b,and c in the given chart.

Answer:
a: Epicarp
b: Mesocarp
c: Endocarp

Question 55.
What are simple fruits? Explain its different types.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato

Question 56.
What are aggregate fruits? Enlist the types of aggregate fruits.
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)

Question 57.
Write a short note on composite fruits.
Answer:
1. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).
2. Fruits which develops from hypanthodium inflorescence are called syconus.

Question 58.
Match the columns with column II.

Column I	Column II
1. Composite fruit	(a) Custard apple
2. Parthenocarpic fruit	(b) Cultivated Banana
3. Aggregate fruit	(c) Mango
	(d) Pineapple

Answer:

Column I	Column II
1. Composite fruit	(d) Pineapple
2. Parthenocarpic fruit	(b) Cultivated Banana
3. Aggregate fruit	(a) Custard apple

Question 59.
Describe the structure of a seed.
Answer:
Structure of a seed:

1. Seed is a reproductive unit that developed from fertilized mature ovule.
2. The seed is made up of seed coat, embryo with or without endosperm and one or two cotyledons.
3. Outer most covering of a seed is called seed coat.
4. It shows outer layer called as testa and inner layer called as tegmen.
5. Hilum is a scar on the seed coat through which seed attach to the fruit.
6. Embryo of a seed is enclosed within seed coat.
7. Embryonal axis consists of radicle and plumule.
8. The part of embryonal axis between cotyledon and plumule is epicotyl, while the part between cotyledons and radicle is hypocotyl.
9. The nutritive tissue in a seed called endosperm.

[Note: Dicotyledonous seed is a non-endospermic or exalbuminous, as it lacks endosperm at maturity.]
[Note: Students can scan the given Q.R code to study the structure of dicotyledonous and monocotyledonous seed.]

Question 60.
Describe the family Fabaceae with suitable floral diagram.
Answer:

• Example: Pea plant (Pisum sativum)
• Habit: Tree, shrubs, herbs.
• Root: Root with root nodules.
• Stem: Erect or climber.
• Leaves: Alternate phyllotaxy, Pinnately compound leaves.
• Inflorescence: Racemose
• Flower: Zygomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, imbricate aestivation.
• Corolla: Petals five, polypetalous, consisting of a larger posterior petal vexillum, two lateral petals wings and two anterior ones forming a keel, vexillary aestivation.
• Androecium: Stamens ten, diadelphous.
• Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation. Fruit: Legume.
• Seed: Non-endospermic

Question 61.
Apply your knowledge

Question 1.
Tendrils are seen in the following plants. Identify whether they are stem tendrils or leaf tendrils,

1. Vitis
2. Smilax
3. Lathyrus
4. Passiflora
5. Cucurbita
6. Gloriosa

Answer:
Stem tendrils → Vitis, Passiflora, Cucurbita
Leaf tendrils → Smilax, Lathyrus, Gloriosa

Question 2.
Which type of roots are shown by halophytes growing in Sundarbans in West Bengal?
Answer:
Pneumatophores

Question 3.
Find out from internet the state flower of Sikkim. Write about the type of roots shown by this plant.
Answer:
Dendrobium is the state flower of Sikkim. It shows epiphytic roots.
Epiphytic roots:

1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question 62.
Quick Review:

Question 63.
Exercise:

Question 1.
Draw neat and labelled diagram of tap root showing different regions.
Answer:
A typical root possesses the following regions:
1. Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root is

2. Meristematic region or region of cell division:
a. The apex of the root is a growing point about 1 mm in length protected by root cap. This region is called as region of cell division or meristematic region.
b. The structure is developed by compactly arranged thin walled actively dividing meristematic cells.
c. These cells bring about longitudinal growth of root.

3. Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

4. Region of root hair or region of absorption:
a. A region of root hair / absorption/piliferous zone is made up of numerous hair like outgrowths.
b. The epiblema or piliferous layer produces tubular elongated unicellular structures known as root hair.
c. They are in close contact with soil particles and increase surface area for absorption of water.
d. Root hair are short lived or ephimeral and are replaced after every 10 to 15 days.

5. Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question 2.
Write a short note on root cap.
Answer:
Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root.

Question 3.
Name the region of a root which is located just above the zone of a cell division.
Answer:
Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

Question 4.
Name any two plants in which root caps are replaced by root pockets.
Answer:
In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.

Question 5.
Write a short note on pneumatophores.
Answer:
1. a. Plants growing in marshy region (halophytes) produce upwardly growing roots called as
pneumatophores or respiratory roots.
b. The main root system of these plants does not get sufficient air for respiration as soil is water logged.
c. Due to this, mineral absorption of plant also gets affected.
d. To overcome this problem underground roots, develop special roots which are negatively geotropic; growing vertically upward.
e. These roots are conical projections present around main trunk of plant.
f. Respiratory roots show presence of lenticels which helps in gaseous exchange.
2. Examples of plants in which respiratory roots are present:
Rhizophora, Avicennia, Sormeratia, Heritiera fames (sundri), etc.

Question 6.
Name the four types of adventitious roots modified for food storage.
Answer:
1. Figure ‘a’ indicates simple tuberous root.
a. Simple tuberous roots become swollen and do not show definite shape.
b. They are produced singly.
c. The roots arise from nodes over the stem and penetrate into the soil, e.g. sweet potato or shakarkand (Ipomoea batatas).

2. Figure ‘b’ indicates fasciculated tuberous roots.
a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

3. Figure ‘c’ indicates Moniliform roots.
a. Some adventitious roots get swollen at regular intervals.
b. These gives them the appearance of beads of a necklace. Such roots are called as Moniliform roots, e.g. Spinacia oleracea (Indian Spinach).

4. Figure ‘d’ indicates Nodulose roots.
The cluster of long slender roots become enlarged at the tips forming nodules is known as nodulose roots, e.g. Arrow root (Maranta), Amhaldi or mango ginger (Curcuma amada).

Question 7.
Explain in detail modification of tap roots for food storage.
Answer:
1. Modification of tap root for storage of food:
a. WTien tap root stores food it becomes swollen, fleshy and also develops definite shape.
b. Main or primary root is the main storage organ but sometimes hypocotyl part of embryo axis also joins the main root. Secondary roots remain thin.
c. On the basis of shape, swollen tap roots are classified as Fusiform, Conical and Napiform.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)
3. Conical root:
The conical root is broad at its morphological base and narrows down towards its apex. e.g. Carrot (Daucus car ota)
4. Napiform root:
In napiform root, base of root is highly swollen, almost spherical in shape and abruptly narrows down towards its apex. e.g. Beet (Beta vulgaris)

Question 8.
Write a short note on:
1. Stilt root
2. Climbing roots
Answer:
1. Stilt roots:
a. These roots normally arise from a few lower nodes of a weak stem in some monocots, shrubs and small trees.
b. They show obliquely downward growth penetrating soil and provide mechanical support to the plant.
c. In the members of family Poaceae, the plants like Maize, Jowar, Sugarcane etc. produce stilt root in whorl around the node.
d. These roots provide additional support to the plant body.
e. In Screw pine or Pandanus (Kewada), stilt roots arise only from the lower surface of obliquely growing stem for additional support. These roots show multiple root caps.

2. Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

Question 9.
What is the function of prop roots in banyan tree?
Answer:
These prop roots show secondary growth, become thick, act like pillars to provide mechanical support to the heavy branches.

Question 10.
Name any two plants which show climbing roots.
Answer:
Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

Question 11.
Identify the type of modified roots shown in the picture given below.

Answer:
Plank roots/Buttress:
a. These roots often develop at the base of large trees and form plank like extensions around stem.
b. These roots provide additional support, e.g. Silk cotton, Peepal, etc.

Question 12.
Identify the type of root marked as ‘X’ in the given picture and write a short note on it.

Answer:
Epiphytic roots:
1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question 13.
Identify the type of aerial modification of stem in the given figure.

Answer:
Cladophylls:
These are leaf like structures bore in the axil of scale leaf. It has floral bud and scale leaf in the middle i.e. upper half is leaf and lower half is stem. e.g. Ruscus.

Question 14.
Write a short note on corm.
Answer:

1. In Colocasia and Amorphophallus corm is present, which is an underground stem modified for storage of food.
2. Corm is swollen underground spherical or subspherical vertically growing stem.
3. It is condensed structure with circular or ring like nodes.
4. It shows presence of axillary buds and scales.
5. Adventitious buds are produced which help in vegetative propagation.
6. Adventitious roots are produced at lower part of the stem.

Question 15.
Observe the type of sub aerial stem in the given picture of Eichhornia and explain it.

Answer:
Offset:
1. These are one intemode long runners in rosette plants at ground or water level.
2. Offset helps in vegetative propagation.
e.g. Water hyacinth or Jal kumbhi (Eichhornia) and Pistia. [Any one example]

Question 16.
Explain the type of stem modification in Opuntia.
Answer:
Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

Question 17.
Identify the structure ‘X’ in the given figure and explain it.

Answer:
Bulbils:
a. In plants like Dioscorea, etc. axillary bud becomes fleshy and rounded due to storage of food called as bulbil.
b. When it falls off it produces new plant and help in vegetative propagation.

Question 18.
Describe the types of stem tendrils present in different plants.
Answer:
1. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.
d. Apical bud in Vitis quadrangularis gets modified into tendril. The further growth is carried out by axillary bud.
e. In Passiflora axillary bud gets modified in tendril.
f. Extra axillary bud is the one which grows outside the axil. This bud in cucurbita gets modified into tendril.
g. Normally floral buds are destined to produce flowers. But in plants like Antigonon they produce tendrils.

Question 19.
What are cladodes?
Answer:
Cladodes:
a. The branches of limited growth i.e. one intemode long and performing photosynthetic function are called as cladodes.
b. True leaves are reduced to spine or scales to reduce rate of transpiration, e.g. Asparagus.

Question 20.
What are tendrils?
Answer:
Tendrils are thin, wiry, photosynthetic, leafless coiled structures.

Question 21.
What is sympodial growth in ginger rhizome and monopodial growth in lotus rhizome?
Answer:
1. Growth of rhizome takes place with lateral buds, such growth is known as sympodial growth, e.g. Ginger (Zingiber officinale), Turmeric (Curcuma domestica), Canna etc.
2. In plants where rhizomes grow obliquely, terminal bud brings about growth of rhizomes. This is known as monopodial growth, e.g. Nymphea, Nelumbo (Lotus), Pteris (Fern) etc.

Question 22.
In xerophytic plant like Opuntia, the stem becomes photosynthetic. Give reason.
Answer:
1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert.
2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration.
3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

Question 23.
Define compound leaves. Explain its two types.
Answer:
The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.
1. Pinnately compound: Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf. e.g. Cassia, Rose, Caesalpinia, Moringa, Coriandrum, etc. [Any two examples]
2. Palmately compound: In this all the leaflets are attached at the tip of petiole.
e.g. Citrus, Zorina, Oxalis, Marsilea, Bombax [Any two examples] [Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 24.
Define leaf venation. What are its two types?
Answer:
Leaf venation:
1. Arrangement of veins and veinlets in leaf lamina is known as venation.
2. There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question 25.
Draw neat and labelled diagram of structure of a typical leaf.
Answer:
A typical dicot leaf shows presence of three main parts Leaf base or Hypopodium, Petiole or Mesopodium and Leaf lamina/blade or Epipodium.
1. Leaf base or Hypopodium:
a. The point by which leaf remains attached to the stem is known as leaf base.
b. The nature of leaf base varies in different plants. It may be pulvinus (swollen), sheathing or ligulate, etc.

Question 26.
Which type of venation can be observed in monocot leaf?
Answer:
There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question 27.
Enlist the types of pinnately compound leaves and give one example of each.
Answer:
Figure ‘a’: Paripinnately compound leaves (e.g. Cassia)
Figure ‘b’: Imparipinnately compound leaves (e.g. Rosa)
Figure ‘c’: Bipinnately compound leaves (e.g. Caesalpinia)
Figure ‘d’: Tripinnately compound leaves (e.g. Moringa)
Figure ‘e’: Decompound leaves (e.g. Coriandrum)

Question 28.
What is opposite decussate phyllotaxy? Give one example.
Answer:
Figure ‘c’ represents opposite decussate phyllotaxy. In this type of phyllotaxy, a pair of leaf arise from each node and the consecutive pair at right angle to the previous one. e.g. Calotropis.

Question 29.
Explain in detail androecium of an angiospermic flower.
Answer:
Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.

2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.

3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.

4. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.

1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 30.
Draw neat and labelled diagram of a typical flower.
Answer:
Floral parts of a typical flower:
1. Calyx (K):
a. It is outermost floral whorl and individual members are known as
sepals.
b. Sepals are usually green in colour and perform photosynthesis.
c. If all the sepals are united, the condition is gamosepalous and if they are free, the condition is called as polysepalous.
d. Gamosepalous calyx is found in China rose and polysepalous calyx is found in Brassica.
e. The main function of sepals is to protect inner floral parts in bud condition.
f. Sometimes sepals become brightly coloured (petaloid sepals) and attract insects for pollination,
e.g. Mussaenda etc.
g. Sepals modify into hairy structures called as pappus. Such calyx helps in dispersal of fruit, e.g. Tridex.

2. Corolla (C):
a. It is second floral whorl from outer side and variously coloured.
b. The individual member is called as petal.
c. Petals may be sweet to taste, possess scent, odour, aroma or fragrance etc.
d. The condition in which petals are free is said to be polypetalous (e.g. Rose) and if they are fused it is called as gamopetalous (e.g. Datura).
e. The main function of corolla is to attract different agencies for pollination.

3. Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

4. Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.
2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.
3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.
4. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.
1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 37.
How many carpels are present in gynoecium of Cucurbita and Hibiscus flower?
Answer:
The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.

Question 38.
Enlist the different types of aestivation and placentation in a flower.
Answer:
Different types of aestivation are as follows:
a. Valvate: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g. Calyx of Datura, Calotropis.
b. Twisted: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g. Corolla of China rose, Cotton etc.
c. Imbricate: One of the sepals or petals is internal and is overlapped at both the margins. One is external i.e. both of its margin overlap adjacent member. Rest of the sepals / petals have one inner or overlapped margin and outer or overlapping margin, e.g. Cassia, Bauhinia, etc.
d. Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

2. Types of placentation:
a. Marginal: Ovules are placed on the fused margins of unilocular ovary, e.g. Pea, Bean etc.
b. Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.
c. Parietal: Ovules are placed on the inner wall of unilocular ovary of multicarpellary syncarpus ovary,
e. g. Papaya, Cucumber, etc.
d. Basal: Single ovule is present at the base of unilocular ovary, e.g. Sunflower, Rice, Wheat.
e. Free central: Ovules are borne on central axis which is not attached to ovary wall, e.g.Argemone, Dianthus.

Question 39.
Explain in detail types of fruits.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
1. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).
2. Fruits which develops from hypanthodium inflorescence are called syconus.

Question 40.
Give one example of each type of fruit given below:
1. Etario of berries
2. Berry
3. Etario of follicles
4. Cypsela
5. Sorosis
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
3. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
4. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)
5. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).

Question 41.
Define parthenocarpic fruit and give one example.
Answer:
Fruits which are produced from ovary without fertilization are called as parthenocarpic fruits, e.g. Cultivated Banana and Grapes.

Question 42.
Write a short note on simple fruits.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato

Question 43.
Give examples of any two types of aggregate fruits.
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)

Question 44.
Complete the given chart.

Answer:
It shows outer layer called as testa and inner layer called as tegmen.

Question 45.
Explain the family of pea plant in detail with suitable floral diagram.
Answer:

• Example: Pea plant (Pisum sativum)
• Habit: Tree, shrubs, herbs.
• Root: Root with root nodules.
• Stem: Erect or climber.
• Leaves: Alternate phyllotaxy, Pinnately compound leaves.
• Inflorescence: Racemose
• Flower: Zygomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, imbricate aestivation.
• Corolla: Petals five, polypetalous, consisting of a larger posterior petal vexillum, two lateral petals wings and two anterior ones forming a keel, vexillary aestivation.
• Androecium: Stamens ten, diadelphous.
• Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation. Fruit: Legume.
• Seed: Non-endospermic

Question 46.
Answer the following.
1. Explain the family Solanaceae with the help of floral diagram.
2. Give examples of economically important plants from family Liliaceae.
Answer:

• Example: Thom apple (Datura stramonium)
• Habit: Mostly herbs, shmbs and rarely small trees.
• Root: Tap root system
• Stem: Herbaceous, woody, erect, branched, hairy, sometimes it may be underground like in potato.
• Leaves: Alternate phyllotaxy, simple, reticulate venation.
• Inflorescence: Solitary, cymose.
• Flower: Actinomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, persistent, valvate aestivation.
• Corolla: Petals five, gamopetalous, valvate aestivation.
• Androecium: Stamens five, free epipetalous (adhesion).
• Gynoecium: Bicarpellary, syncarpous, superior ovary, bilocular, placenta swollen with many ovules, axile placentation.
• Fruits: Berry or capsule.
• Seeds: Many, endospermic.

2. Economically important plant from family Liliaceae:
Family Liliaceae includes many ornamental plants like tulip, Gloriosa, Medicinal plants like Aloe vera. Asparagus and source of colchicine, e.g. Colchicum autumnale.

Question 64.
Multiple Choice Questions:

Question 1.
Root is descending axis of plant body which is
(A) negatively geotropic
(B) hydrophobic
(C) negatively phototropic
(D) green and with intemodes
Answer:
(C) negatively phototropic

Question 2.
The root system grow out from the
(A) plumule of the embryo
(B) radicle of the embryo
(C) embryo of the seed
(D) all of these
Answer:
(B) radicle of the embryo

Question 3.
Adventitious roots develop from
(A) radicle
(B) any part of the plant body except the radicle
(C) flower
(D) embryo
Answer:
(B) any part of the plant body except the radicle

Question 4.
A fibrous root system is best adapted to perform which of the following functions?
(A) Storage of food
(B) Transport of water and organic food
(C) Absorption of water and minerals from the soil
(D) Anchorage of the plant into the soil
Answer:
(D) Anchorage of the plant into the soil

Question 5.
When the root is swollen in the middle and tapers at both ends, it will be called as root.
(A) tuberous
(B) fusiform
(C) conical
(D) napiform
Answer:
(B) fusiform

Question 6.
Pneumatophores are helpful in
(A) protein synthesis
(B) respiration
(C) transpiration
(D) carbohydrate metabolism
Answer:
(B) respiration

Question 7.
Sweet potato is a modification of
(A) leaf
(B) adventitious root
(C) tap root
(D) stem
Answer:
(B) adventitious root

Question 8.
Stilt roots are roots.
(A) primary
(B) adventitious
(C) secondary
(D) tap
Answer:
(B) adventitious

Question 9.
A spongy tissue called velamen is present in
(A) breathing roots
(B) parasitic roots
(C) tuberous roots
(D) epiphytic roots
Answer:
(D) epiphytic roots

Question 10.
Which of the following is NOT a type of adventitious root modified for storage of food?
(A) Fasciculated tuberous roots
(B) Simple tuberous root
(C) Napiform root
(D) Moniliform roots
Answer:
(C) Napiform root

Question 11.
In which of the following plants, root cap is replaced by root pocket?
(A) Pistia
(B) Pandanus
(C) Screw pine
(D) Hibiscus
Answer:
(A) Pistia

Question 12.
Which of the following is an example of stem tuber?
(A) Helianthus tuberosus
(B) Zingiber officinale
(C) Cynodon
(D) Chrysanthemum
Answer:
(A) Helianthus tuberosus

Question 13.
In stem tuber, the number of nodes and eyes is more towards
(A) rose end
(B) basal end
(C) heel
(D) both (B) and (C)
Answer:
(A) rose end

Question 14.
A rhizome differs from corm in its
(A) thickness
(B) basic organization
(C) direction of growth
(D) nature of leaves
Answer:
(C) direction of growth

Question 15.
The reduced stem of onion produces.
(A) Adventitious roots
(B) Prop roots
(C) Fusiform roots
(D) Fasciculated roots
Answer:
(A) Adventitious roots

Question 16.
Corm is ____________
(A) a horizontal underground stem.
(B) an underground root.
(C) an underground vertical stem.
(D) an aerial stem modification.
Answer:
(C) an underground vertical stem.

Question 17.
The stem modified to perform the function of leaf and with many internodes is called
(A) phylloclade
(B) cladode
(C) offset
(D) phyllode
Answer:
(A) phylloclade

Question 18.
_______ is a non-green runner like branch of stem, which develops from underground base of roots and found in Chrysanthemum.
(A) Corn
(B) Sucker
(C) Offset
(D) Tendril
Answer:
(B) Sucker

Question 19.
Ribbon shaped phylloclades are found in
(A) Ruscus
(B) Duranta
(C) Muehlenbeckia
(D) Bougainvillea
Answer:
(C) Muehlenbeckia

Question 20.
Axillary buds in Dioscorea becomes fleshy and rounded due to storage of food called as
(A) Stiphles
(B) Bulbils
(C) Offset
(D) Cladophylls
Answer:
(B) Bulbils

Question 21.
The leaves without petiole are called
(A) sessile
(B) petiolate
(C) rachis
(D) lamina
Answer:
(A) sessile

Question 22.
The type of leaves observed is mango plant is
(A) Compound leaves
(B) Bipinnately compound leaves
(C) Simple leaves with reticulate venation
(D) Simple leaves with parallel venation
Answer:
(C) Simple leaves with reticulate venation

Question 23.
In _________ , the terminal three leaflet get modified into three stiff leaf hooks.
(A) Lathyrus
(B) Pisum sativum
(C) Smilax
(D) Bignonia unguisi-cati
Answer:
(D) Bignonia unguisi-cati

Question 24.
Leaf apex is modified into tendril in
(A) Gloriosa
(B) Pea
(C) Smilax
(D) Lathyrus
Answer:
(A) Gloriosa

Question 25.
Modification of petiole into leaf-like structure is called ________ .
(A) cladode
(B) phylloclade
(C) phyllode
(D) pistillode
Answer:
(C) phyllode

Question 26.
The mode of arrangement of leaves on the stem and the branch is known as _______ .
(A) vernalization
(B) vernation
(C) venation
(D) phyllotaxy
Answer:
(D) phyllotaxy

Question 27.
Bipinnately compound leaves can be observed in
(A) Citrus
(B) Hibiscus
(C) Caesalpinia
(D) Coriandrum
Answer:
(C) Caesalpinia

Question 28.
The axis of the inflorescence is known as
(A) Thalamus
(B) Peduncle
(C) Pedicel
(D) Petiole
Answer:
(B) Peduncle

Question 29.
In racemose inflorescence
(A) growth of peduncle is infinite
(B) apical bud always terminates into flower
(C) growth of peduncle is finite.
(D) order of opening of flower is centrifugal.
Answer:
(A) growth of peduncle is infinite

Question 30.
In a typical flower thalamus consists of compactly arranged nodes and intemodes.
(A) three, two
(B) four, three
(C) three, four
(D) two, three
Answer:
(B) four, three

Question 31.
When the flower is epigynous, the ovary is said to be
(A) inferior
(B) superior
(C) semi-inferior
(D) semi-superior
Answer:
(A) inferior

Question 32.
When the gynoecium is present at the topmost position of the thalamus, the flower is known as
(A) inferior
(B) epigynous
(C) perigynous
(D) hypogynous
Answer:
(D) hypogynous

Question 33.
When all sepals are united, the condition is called as
(A) polysepalous
(B) gamosepalous
(C) polypetalous
(D) gamopetalous
Answer:
(B) gamosepalous

Question 34.
When sepals fall just after opening of the flower, they are termed as
(A) persistent
(B) caducous
(C) remnant
(D) deciduous
Answer:
(B) caducous

Question 35.
Fusion between members of a similar whorl is known as
(A) succession
(B) adhesion
(C) cohesion
(D) inflorescence
Answer:
(C) cohesion

Question 36.
Complete the analogy. Seed:ovule::Fruit:
(A) pericarp
(B) ovary
(C) embryo
(D) cotyledons
Answer:
(B) ovary

Question 37.
Which one of the following is not a fruit?
(A) Tomato
(B) Cucumber
(C) Pumpkin
(D) Potato
Answer:
(D) Potato

Question 38.
Pineapple is an example of _________ .
(A) simple dry fruit
(B) composite fruit
(C) aggregate fruit
(D) simple-fleshy fruit
Answer:
(B) composite fruit

Question 39.
Outer seed coat is called _______ .
(A) testa
(B) tegmen
(C) raphe
(D) micropyle
Answer:
(A) testa

Question 40.
Vexillum wings and keel corolla are found in family
(A) Solanaceae
(B) Fabaceae
(C) Liliaceae
(D) Malvaceae
Answer:
(B) Fabaceae

Question 65.
Competitive Corner:

Question 1.
Match the placental types (Column-I) with their examples (Column-II).

Column-I	Column-II
1. Basal	(p) Mustard
2. Axile	(q) China rose
3. Parietal	(r) Dianthus
4. Free central	(s) Sunflower

Choose the correct answer from the following option: [NEET (ODISHA) – 2019J
(A) i – r, ii – s, iii – p, iv – q
(B) i – q, ii – r, iii – s, iv – p
(C) i – p, ii – q, iii – r, iv – s
(D) i – s, ii – q, iii – p, iv – r
Answer:
(D) i – s, ii – q, iii – p, iv – r

Question 2.
Placentation in which ovules develop on the inner wall of the ovary or in peripheral part is:
(A) Parietal
(B) Free central
(C) Basal
(D) Axile
Answer:
(A) Parietal

Question 3.
Pneumatophores occur in
(A) Carnivorous plants
(B) Free-floating hydrophytes
(C) Halophytes
(D) Submerged hydrophytes
Answer:
(C) Halophytes

Question 4.
Sweet potato is a modified
(A) Tap root
(B) Adventitious root
(C) Stem
(D) Rhizome
Answer:
(B) Adventitious root

Question 5.
Root hairs develop from the region of
(A) Maturation
(B) Elongation
(C) Root cap
(D) Meristematic activity
Hint: Epidermal cells from the region of maturation form very fine and delicate, thread like structures called root hairs. These root hairs absorb water and minerals from the soil.
Answer:
(A) Maturation

Question 6.
Plants which produce characteristic pneumatophores and show vivipary belong to
(A) Mesophytes
(B) Halophytes
(C) Psammophytes
(D) Elydrophytes
Hint: Plants growing in swampy areas, marshy places and salt lakes are called halophytes. Many halophytes develop respiratory roots or pneumatophores. Pneumatophores are negatively geotropic and are provided with pores called lenticels. Since they grow in oxygen-deficient soil, their seeds germinate inside the fruit, when it is still attached with the parents, exhibiting vivipary.
Answer:
(B) Halophytes

Question 7.
In Bougainvillea thorns are the modifications of
(A) Stipules
(B) Adventitious root
(C) Stem
(D) Leaf
Hint: Thom is a hard, pointed, woody and usually straight structure produced by modification of axillary bud. It provides protection against browsing animals. In Bougainvillea, thorns are modified stems.
Answer:
(C) Stem

Question 8.
Coconut fruit is a
(A) Drupe
(B) Berry
(C) Nut
(D) Capsule
Hint: Dmpe is the simple, fleshy fruit developing from the monocarpellary superior ovary. It is generally one seeded. Coconut fruit is a dmpe with fibrous mesocarp and hard endocarp
Answer:
(A) Drupe

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 6 Bank Reconciliation Statement Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 6 Bank Reconciliation Statement

1. Answer in one sentence.

Question 1.
What is a Bank Reconciliation Statement?
Answer:
Bank Reconciliation Statement is a statement that shows the causes of disagreement between the balance shown by passbook and the balance shown by Cash Book under the column as on a particular date.

Question 2.
What is a Bank passbook?
Answer:
A Bank passbook is a copy of a Customer’s A/c in the Bank’s ledger.

Question 3.
What do you mean by the debit balance of Pass Book?
Answer:
Debit balance of passbook means overdraft as per passbook.

Question 4.
Which account is opened by a trader in Bank for his business operation?
Answer:
A current account is opened by a trader in Bank for his business operation.

Question 5.
On which side of the Cash Book interest on investment is to be shown?
Answer:
Interest on investment is to be shown on the debit/receipt side in the Cash Book.

Question 6.
On which side of the passbook, the direct deposit made by a customer is recorded?
Answer:
A direct deposit made by a customer is recorded on the credit side of the passbook.

Question 7.
What does the credit balance of Cash Book indicate?
Answer:
A credit balance of Cash Book indicates overdraft as per Cash Book.

2. Give one word/term/phrase which can substitute each of the following statements:

Question 1.
Excess of a total of debit side over a total of credit side of Cash Book, Bank column.
Answer:
Bank Balance as per Cash Book

Question 2.
An unfavourable balance is shown by the passbook.
Answer:
Overdraft

Question 3.
A copy of the customer’s account issued by the bank.
Answer:
Pass Book

Question 4.
Booklet or a statement that is used to record the banking transactions.
Answer:
Pass Book

Question 5.
The credit balance of the bank column of the Cash Book.
Answer:
Overdraft

Question 6.
Refusal by the bank to make payment of a cheque.
Answer:
Dishonour of Cheque

Question 7.
Document used to withdraw cash from the bank.
Answer:
Withdrawal Slip

Question 8.
Excess of a total of credit side over a total of debit side in the passbook.
Answer:
Balance as per Pass Book

Question 9.
Document used by the account holder to deposit cash and/or cheques into the bank.
Answer:
Pay-in-Slip

Question 10.
Document used by the account holder to withdraw cash from the bank and for making payment to outside parties through the bank.
Answer:
Cheque

3. Do you agree or disagree with the following statements:

Question 1.
Bank Reconciliation Statement is prepared by Bank.
Answer:
Disagree

Question 2.
Overdraft Facility is available to savings Bank Account.
Answer:
Disagree

Question 3.
The bank account holder can make payments to a third party by use of a pay-in slip.
Answer:
Disagree

Question 4.
Bank does not charge any interest on an overdraft balance.
Answer:
Disagree

Question 5.
Credit balance in Pass Book represents overdraft balance.
Answer:
Disagree

Question 6.
In the cash book of a trader Bank record the entries.
Answer:
Disagree

Question 7.
In the passbook, entries are made by the account holder.
Answer:
Disagree

Question 8.
Through mobile banking account holder can deposit physical cash into his account.
Answer:
Disagree

Question 9.
The right-hand side of the pay-in-slip is known as a counterfoil.
Answer:
Disagree

Question 10.
A cash withdrawal slip is used to deposit a cheque or cash into a bank account.
Answer:
Disagree

4. Select the most appropriate alternative from those given and rewrite the following statements:

Question 1.
Pass Book is _____________ of account holders transactions with bank.
(a) an extract
(b) balance sheet
(c) balance
(d) mode
Answer:
(a) an extract

Question 2.
When cheque is _____________ into bank Cash Book is debited.
(a) written
(b) issued
(c) deposited
(d) dishonoured
Answer:
(c) deposited

Question 3.
Overdraft facility is allowed to _____________ account.
(a) saving
(b) recurring
(c) current
(d) fixed deposit.
Answer:
(c) current

Question 4.
Overdraft means _____________ balance of Pass Book.
(a) opening
(b) debit
(c) credit
(d) closing
Answer:
(b) debit

Question 5.
Interest on bank overdraft is recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) any
(d) both
Answer:
(a) debit

Question 6.
Credit balance in the Pass Book represents _____________
(a) overdraft
(b) bank balance
(c) loan borrowed
(d) negative
Answer:
(b) bank balance

Question 7.
Direct deposit by a customer will be recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) left hand
(d) any
Answer:
(b) credit

Question 8.
Normally the Bank Reconciliation Statement is prepared at the end of a _____________
(a) day
(b) week
(c) year
(d) month
Answer:
(d) month

Question 9.
Bank charges charged by bank are recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) any
(d) both
Answer:
(a) debit

5. Complete the following statements:

Question 1.
Bank issue _____________ to current account holders as record or summary for bank transactions.
Answer:
Bank statement

Question 2.
To deposit cash or cheque _____________ is used.
Answer:
Pay in Slip

Question 3.
Directly deposited by a customer in bank account appears an _____________ side of pass book.
Answer:
Credit

Question 4.
____________ is an unfavorable balance as per Pass Book.
Answer:
overdraft

Question 5.
Check deposited into Bank but not cleared is called _____________ cheque.
Answer:
Dishonoured

Question 6.
Expenses paid by the Bank as per standing instructions will be recorded at _____________ side of the passbook.
Answer:
Debit

Question 7.
Dividend collected by the bank will be recorded at _____________ side of cash book.
Answer:
Debit

Question 8.
_____________ type of bank account is open by trader.
Answer:
Current

6. State whether the following statements are True or False with reasons:

Question 1.
Bank Reconciliation Statement is prepared by the Bank.
Answer:
This statement is False.
A Bank Reconciliation statement shows the causes of disagreement between the balances shown by the bank passbook and the bank balance shown by the cash book, for a particular period of time generally a month. It is prepared by the trader as he has both the books to compare and find the differences.

Question 2.
Bank Reconciliation Statement is prepared at the end of every month.
Answer:
This statement is True.
Monthly preparation of Bank Reconciliation Statement assists in the regular monitoring of cash flows of a business and identification of accounting errors.

Question 3.
Overdraft facility is allowed to Proprietor’s Personal A/c.
Answer:
This statement is False.
Overdraft facility is allowed only to business current A/c and not to Proprietor’s Personal A/c.

Question 4.
The debit balance of Pass Book represents overdraft.
Answer:
This statement is True.
Debit Column of a passbook means withdrawals from the bank, When withdrawals are more than deposits it means the excess amount is withdrawn from the bank. It is a temporary loan payable by the trader to the bank. So the debit balance of Pass Book represents overdraft.

Question 6.
Bank charges debited by Bank increase bank balance as per Pass Book.
Answer:
This statement is False.
Bank charges are expenses for the business. Expenses decrease the bank balance as per Pass Book. Bank charges debited by the bank decrease the bank balance as per Pass Book.

Question 6.
Interest credited in PassBook is an income to the customer.
Answer:
This statement is True.
All incomes are shown on the credit side of the passbook. It is a deposit. So Interest Credited in Passbook is an income to the customer.

Question 7.
Bank Reconciliation Statement is prepared to detect the errors that take place in accounting.
Answer:
This statement is True.
A businessman maintains a cash book with a bank column to record his bank transactions whereas the bank also maintains a customer’s ledger account and issues him a Bank statement. There could be differences as per the bank balance in the cash book and bank balance in the passbook. To detect the errors that take place in accounting Bank Reconciliation Statement is prepared.

Question 8.
Overdraft as per Cash Book means debit balance as per Cash Book.
Answer:
This statement is False.
Overdraft as per Cash Book means credit balance as per cash book. Cashbook debit means deposits. When cash book debit balance is greater it means Bank Balance as per Cash Book.

Question 9.
Cheque deposited into Bank increases the Bank balance as per Cash Book.
Answer:
This statement is True.
The Debit side of the cash book means deposits. So cheque deposited into the Bank increases the Bank Balance as per Cash Book.

Question 10.
Payments made by the bank as per standing instructions are recorded on the Debit balance of the Pass Book.
Answer:
This statement is True.
The Debit side of the passbook represents payments. So any payments made by the bank, Bank debits the customer’s account and records on the Debit side of the passbook.

Question 11.
Bank column in Cash Book represents Proprietor’s Savings A/c.
Answer:
This statement is False.
Cashbook records only business transactions and not the personal A/c of the trader. The Bank column in Cash Book represents Business Current A/c and not the proprietor’s savings A/c.

7. Correct and rewrite the following statements.

Question 1.
Overdraft as per cash book means debit balance as per cash book.
Answer:
Normal bank balance as per cash book means debit balance as per cashback.

Question 2.
Bank column in cash book represents proprietors saving A/c.
Answer:
The Bank column in cashback represents the business’s current account.

Question 3.
Fixed deposit A/c is opened by traders for day-to-day business bank transactions.
Answer:
The current account is opened by traders for day-to-day business bank transactions.

Question 4.
The bank account is a real account.
Answer:
A bank account is a personal account

Question 5.
Interest charged by the bank on overdraft A/c is income for the business.
Answer:
Interest charged by the bank on overdraft A/c is an expense for the business.

8. Complete the following table.

Question 1.


Answer:

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 2 Introduction to Analytical Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 2 Introduction to Analytical Chemistry

Question 1.
Explain the statement: Analytical chemistry provides physical or chemical information about a sample.
Answer:

• Analytical chemistry facilitates the investigation of the chemical composition of substances.
• It uses the instruments and methods to separate, identify and quantify a sample under study.

Thus, analytical chemistry provides chemical or physical information about a sample.

Question 2.
What is the difference between qualitative analysis and quantitative analysis?
Answer:

1. Qualitative analysis deals with the detection of the presence or absence of elements in compounds and of chemical compounds in mixtures.
2. Quantitative analysis deals with the determination of the relative proportions of elements in compounds and of chemical compounds in mixtures.

Question 3.
Explain the importance of chemical analysis.
Answer:

• Chemical analysis is one of the most important methods of monitoring the composition of raw materials, intermediates and finished products, and also the composition of air in streets and premises of industrial plants.
• In agriculture, chemical analysis is used to determine the composition of soils and fertilizers.
• In medicine, it is used to determine the composition of medicinal preparations.

Question 4.
Write a note on the applications of analytical chemistry.
Answer:

• Analytical chemistry has applications in forensic science, engineering and industry.
• Analytical chemistry is also useful in the field of agriculture and pharmaceutical industry.
• Industrial process as a whole and the production of new kinds of materials are closely associated with analytical chemistry.

Question 5.
What is semi-microanalysis?
Answer:

• When the amount of a solid or liquid sample taken for analysis is a few grams, the analysis is called semi-microanalysis.
• It is of two types: qualitative and quantitative analysis.

Question 6.
What does classical qualitative analysis method include?
Answer:
Classical qualitative analysis method includes separation and identification of compounds.

• Separations may be done by methods such as precipitation, extraction and distillation.
• Identification may be based on differences in colour, odour, melting point, boiling point, and reactivity.

Question 7.
Name two methods of classical quantitative analysis.
Answer:

1. Volumetric analysis (Titrimetric analysis)
2. Gravimetric analysis (i.e., decomposition, precipitation)

Question 8.
What are the two stages involved in the chemical analysis of a sample?
Answer:
The chemical analysis of a sample is carried out mainly in two stages: by the dry method and by the wet method. In dry method, the sample under test is not dissolved and in wet method, the sample under test is first dissolved and then analysed to determine its composition.

Question 9.
Explain: Qualitative analysis of organic compounds
Answer:

• The majority of organic compounds are composed of a relatively small number of elements.
• The most important ones are: carbon, hydrogen, oxygen, nitrogen, sulphur, halogen, phosphorus.
• Elementary qualitative analysis is concerned with the detection of the presence of these elements.
• The identification of an organic compound involves tests such as detection of functional group, determination of melting/boiling points, etc.

Question 10.
What does qualitative analysis of inorganic compounds involve?
Answer:
The qualitative analysis of simple inorganic compounds involves detection and confirmation of cationic and anionic species (basic and acidic radical) in them.

Question 11.
Explain: Chemical methods of quantitative analysis
Answer:

• Quantitative analysis of organic compounds involves methods such as determination of percentage of constituent elements, concentrations of a known compound in the given sample, etc.
• Quantitative analysis of simple inorganic compounds involves methods such as gravimetric analysis (i.e., decomposition, precipitation, etc.) and the titrimetric or volumetric analysis (i.e., progress of reaction between two solutions till its completion).
• The quantitative analytical methods involve measurement of quantities such as mass and volume using some equipment/apparatus such as weighing machine, burette, etc.

Question 12.
Why is accurate measurement crucial in science?
Answer:

• The accuracy of measurement is of a great concern in analytical chemistry. This is because faulty equipment, poor data processing or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in the analytical measurement.
• When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.
• Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
• Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation and properly express the quantitative error in the result.

Question 13.
Why are scientific notations (exponential notations) used?
Answer:
A chemist has to deal with numbers as large as 602,200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g. that is, mass of a H atom. To avoid the writing of so many zeros in mathematical operations, scientific notations i.e. exponential notations are used.

Question 14.
How are numbers expressed in scientific notations (exponential notations)?
Answer:
In scientific notations, numbers are expressed in the form of N × 10ⁿ, where ‘n’ is an exponent with positive or negative values and N can have a value between 1 to 10.
e.g. i. The number, 602,200,000,000,000,000,000,000 is expressed as 6.022 × 10²³.
ii. The mass of a H atom, 0.00000000000000000000000166 g is expressed as 1.66 × 10^-24 g.
iii. The number 123.546 is written as 1.23546 × 10².
iv. The number 0.00015 is written as 1.5 × 10^-4.

Question 15.
Express the following quantities in scientific notations (exponential notations).
i. 0.0345
ii . 0.08
iii. 653.00
iv. 34.768
Answer:
i. 0.0345 = 3.45 × 10^-2
ii. 0.08 = 8 × 10^-2
iii. 653.00 = 6.5300 × 10²
iv. 34.768 = 3.4768 × 10¹

Question 16.
Define: Accuracy of measurement
Answer:
Nearness of the measured value to the true value is called the accuracy of measurement.

Question 17.
Explain with the help of a diagram how accuracy depends upon the sensitivity or least count of the measuring equipment.
Answer:
A burette reading of 10.2 mL is as shown in the diagram below:

• For all the three situations in the above figure, the reading would be noted is 10.2 mL.
• It means that there is an uncertainty about the digit appearing after the decimal point in the reading 10.2 mL because the least count of the burette is 0.1 mL.
• The meaning of the reading 10.2 mL is that the true value of the reading lies between 10.1 mL and 10.3 mL.
• This is indicated by writing 10.2 ± 0.1 mL.
• Here, the burette reading has an error of ± 0.1 mL.
• Smaller the error, higher is the accuracy.

Question 18.
How is absolute error calculated?
Answer:
Absolute error is calculated by subtracting true value from observed value.
Absolute error = Observed value – Tme value

Question 19.
Explain the term: Relative error
Answer:

1. Relative error is the ratio of an absolute error to the true value.
2. Relative error is generally a more useful quantity than absolute error.
3. Relative error is expressed as a percentage and can be calculated as follows:
   Relative error = \(\frac{\text { Absolute error }}{\text { True value }} \times 100 \%\)

Question 20.
Explain the term: Precision in measurement
Answer:

• Multiple readings of the same quantity are noted to minimize the error.
• If the multiple readings of the same quantity match closely, they are said to have high precision.
• High precision implies reproducibility of the readings.
• High precision is a prerequisite for high accuracy.
• Precision is expressed in terms of deviation (i.e. absolute deviation and relative deviation).

Question 21.
Explain the following terms with respect to precise measurement:
i. Absolute deviation
ii. Mean absolute deviation
iii. Relative deviation
Answer:
i. Absolute deviation: An absolute deviation is the modulus of the difference between an observed value and the arithmetic mean for the set of several measurements made in the same way. It is a measure of absolute error in the repeated observation. It is expressed as follows:
Absolute deviation = |Observed value – Mean|

ii. Mean absolute deviation: Arithmetic mean of all the absolute deviations is called the mean absolute deviation in the measurements.

iii. Relative deviation: The ratio of mean absolute deviation to its arithmetic mean is called relative deviation. It is expressed as follows:
Relative deviation = \(\frac{\text { Mean absolute deviation }}{\text { Mean }}\) × 100%

Question 22.
Explain the need of significant figures in measurement.
Answer:

• Uncertainty in measured value leads to uncertainty in calculated result.
• Uncertainty in a value is indicated by mentioning the number of significant figures in that value. e.g. Consider, the column reading 10.2 ± 0.1 mL recorded on a burette having the least count of 0.1 mL. Here, it is said that the last digit ‘2’ in the reading is uncertain, its uncertainty is ±0.1 mL. On the other hand, the figure ‘10’ is certain.
• The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
• In a scientific experiment, a result is obtained by doing calculation in which values of a number of quantities measured with equipment of different least counts are used.

Question 23.
How many significant figures are present in the following measurements?
i. 4.065 m
ii. 0.32 g
iii. 57.98 cm³
iv. 0.02 s
v. 4.0 × 10^-4 km
vi. 604.0820 kg
vii. 307.100 × 10^-5 cm
Answer:
i. 4
ii. 2
iii. 4
iv. 1
v. 2
vi. 7
vii. 6

Question 24.
How many significant figures are present in each of the following?
i. 45.0
ii. 0.001
iii. 2.10 × 10^-8
iv. 340000
v. 0.0100
vi. 7890320
vii. 100.00
viii. 100
Answer:
i. 3
ii. 1
iii. 3
iv. 2
v. 3
vi. 6
vii. 5
viii. 1

Question 25.
State the rules used to round off a number to the required number of significant figures.
Answer:
The following rules are used to round off a number to the required number of significant figures:

• If the digit following the last digit to be kept is less than five, the last digit is left unchanged, e.g. 46.32 rounded off to two significant figures is 46.
• If the digit following the last digit to be kept is five or more, the last digit to be kept is increased by one. e.g. 52.87 rounded to three significant figures is 52.9.

Question 26.
Round off each of the following to the number of significant digits indicated:
i. 1.223 to two digits
ii. 12.56 to three digits
iii. 122.17 to four digits
iv. 231.5 to three digits
Answer:
i. 1.223 to two digits = 1.2
This is because the third digit is less than 5, so we drop it and all the other digits to its right.
ii. 12.56 to three digits = 12.6
This is because the fourth digit is greater than 5, so we drop it and add 1 to the third digit.
iii. 122.17 to four digits = 122.2
This is because the fifth digit is greater than 5, so we drop it and add 1 to the fourth digit.
iv. 231.5 to three digits = 232
This is because the fourth digit is 5, so we drop it and add 1 to the third digit.

Question 27.
Add 5.55 × 10⁴ and 6.95 × 10³ and express the result in scientific notation.
Solution:
To perform addition operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added.
(5.55 × 10⁴) + (6.95 × 10³)= (5.55 × 10⁴) + (0.695 × 10⁴)
= (5.55 +0.695) × 10⁴
= 6.245 × 10⁴

Question 28.
Add 1.77 × 10² and 2.23 × 10³ and express the result in scientific notation.
Solution:
To perform addition operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added.
(1.77 × 10²) + (2.23 × 10³) = (0.177 × 10³) + (2.23 × 10³)
= (0.177 + 2.23) × 10³
= 2.407 × 10³

Question 29.
Subtract 5.8 × 10^-3 from 3.5 × 10^-2 and express result in scientific notation.
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
(3.5 × 10^-2)- (5.8 × 10^-3) = (3.5 × 10^-2) – (0.58 × 10^-2)
= (3.5 – 0.58) × 10^-2
= 2.92 × 10^-2

Question 30.
Subtract 6.90 × 10^-5 from 5.11 × 10^-4 and express the result in scientific notation.
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
(5.11 × 10^-4) – (6.90 × 10^-5) = (5.11 × 10^-4) – (0.690 × 10^-4)
= (5.11 – 0.690) × 10^-4
= 4.42 × 10^-4

Question 31.
Perform following calculations and express results in scientific notations (exponential notations),
i. (1.5 × 10^-6) – (5.8 × 10^-7)
ii. (9.8 × 10^-3) – (8.8 × 10^-3)
iii. (6.5 × 10^-8) – (5.5 × 10^-9)
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
i. (1.5 × 10^-6) – (5.8 × 10^-7) = (1.5 × 10^-6) – (0.58 × 10^-6)
= (1.5 – 0.58) × 10^-6
= 0.92 × 10^-6
= 9.2 × 10^-7

ii. (9.8 × 10^-3) – (8.8 × 10^-3) = (9.8 – 8.8) × 10^-3
= 1.0 × 10^-3

iii. (6.5 × 10^-8) – (5.0 × 10^-9) = (6.5 × 10^-8) – (0.50 × 10^-8)
= (6.5 – 0.50) × 10^-8
= 6.0 × 10^-8

Question 32.
Multiply 5.6 × 10⁵ and 6.9 × 10⁸ and express result in scientific notation.
Solution:
(5.6 × 10⁵) × (6.9 × 10⁸) = (5.6 × 6.9) (10⁵⁺⁸)
= 38.64 × 10¹³
= 3.864 × 10¹⁴

Question 33.
Multiply 9.8 × 10^-2 and 2.5 × 10^-6 and express result in scientific notation.
Solution:
(9.8 × 10^-2) × (2.5 × 10^-6) = (9.8 × 2.5) (10^-2+(-6))
= (9.8 × 2.5) × (10^-2-6)
= 24.5 × 10^-8
= 2.45 × 10^-7

Question 34.
Perform following calculations and express results in scientific notations (exponential notations).
i. (2.5 × 10^-6) × (1.8 × 10^-7)
ii. (4.5 × 10^-3) × (1.8 × 10³)
iii. (8.5 × 10⁷) × (3.5 × 10⁹)
Solution:
i. (2.5 × 10^-6) × (1.8 × 10^-7) = (2.5 × 1.8) (10^-6+(-7))
= (2.5 × 1.8) × (10^-6-7)
= 4.5 × 10^-13

ii. (4.5 × 10-3) × (1.8 × 103) = (4.5 × 1.8) (10^-3+3)
= (4.5 × 1.8) × (10⁰)
= 8.1

iii. (8.5 × 10⁷) × (3.5 × 10⁹) = (8.5 × 3.5) (10⁷⁺⁹)
= (8.5 × 3.5) × 10¹⁶
= 29.75 × 10¹⁶
= 2.975 × 10¹⁷
[Note: To express number in scientific notation, the number has to be greater than or equal to 10 or less than 1. The number, 8.1 is greater than 1, but less than 10 and hence, it cannot be expressed in scientific notation.]

Question 35.
In laboratory experiment, 10 g potassium chlorate sample on decomposition gives following data: The sample contains 3.8 g of oxygen and the actual mass of oxygen in the quantity of potassium chlorate is 3.92 g. Calculate absolute error and relative error.
Solution:
The observed value is 3.8 g and accepted (true) value is 3.92 g.
i. Absolute error = Observed value – True value
= 3.8 – 3.92
= -0.12 g

[Note: The negative sign indicates that experimental result is lower than the true value.]
Ans: i. Absolute error = -0.12 g
ii. Relative error = -3.06%

Question 36.
12.15 g of magnesium gives 20.20 g of magnesium oxide on burning. The actual mass of magnesium oxide that should be produced is 20.15 g. Calculate absolute error and relative error.
Solution:
The observed value is 20.20 g and accepted (true) value is 20.15 g.
i. Absolute error = Observed value – True value
= 20.20 – 20.15
= 0.05 g

Ans: i. Absolute error = 0.05 g
ii. Relative error = 0.25 %

Question 37.
The three identical samples of potassium chlorate are decomposed. The mass of oxygen is determined to be 3.87 g, 3.95 g and 3.89 g for the set. Calculate absolute deviation and relative deviation.
Solution:
Mean = \(\frac{3.87+3.95+3.89}{3}\) = 3.90

Sample	Mass of oxygen	Absolute deviation = | Observed value – Mean |
1	3.87 g	0.03 g
2	3.95 g	0.05 g
3	3.89 g	0.01 g

Ans: i. Absolute deviation in each observation = 0.03 g, 0.05 g, 0.01 g
Relative deviation = 0.8%

Question 38.
In repeated measurements in volumetric analysis, the end-points were observed as 11.15 mL, 11.17 mL, 11.11 mL and 11.17 mL. Calculate mean absolute deviation and relative deviation.
Solution:
Mean = \(\frac{11.15+11.17+11.11+11.17}{4}\) = 11.15

Measurement	End-point	Absolute deviation = |Observed value – Mean|
1	11.15 mL	0
2	11.17 mL	0.02 mL
3	11.11 mL	0.04 mL
4	11.17 mL	0.02 mL

Ans: i. Mean absolute deviation = ±0.02 mL
ii. Relative deviation = 0.2%

Question 39.
Calculate the mass percentages of H, P and O in phosphoric acid if atomic masses are H = 1, P = 31 and O = 16.
Solution:
Atomic mass of H = 1, P = 31 and 0=16
The mass percentage of hydrogen, phosphorus, oxygen in H₃PO₄
Formula:

Calculation: Molecular formula of phosphoric acid: H₃PO₄
∴ Molar mass of H₃PO₄ = 3 × (1) + 1 × (31) + 4 × (16)
= 3 + 31 + 64
= 98 g mol^-1
Percentage of H = \(\frac {3}{98}\) × 100 = 3.06%
Percentage of P = \(\frac {31}{98}\) × 100 = 31.63%
Percentage of O = \(\frac {64}{98}\) × 100 = 65.31%
Ans: Mass percentage of H, P and O in phosphoric acid are 3.06%, 31.63% and 65.31% respectively.

Question 40.
Calculate the percentage composition of the elements in HNO₃ (H = 1, N = 14, O = 16).
Solution:
Atomic mass of H = 1, N = 14 and O = 16
The mass percentage of H, N and O in HNO₃
Formula:

Calculation: Molecular formula of HNO₃
∴ Molar mass = 1 × (1) + 1 × (14) + 3 × (16) = 1 + 14 + 48 = 63 g mol^-1
∴ Percentage of H = \(\frac {1}{63}\) × 100 = 1.59%
Percentage of N = \(\frac {14}{63}\) × 100 = 22.22%
Percentage of O = \(\frac {48}{63}\) × 100 = 76.19%
Ans: Mass percentage of H, N and O in HNO₃ are 1.59%, 22.22% and 76.19% respectively.

Question 41.
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass. Its molar mass is 98.96 g mol^-1. What is its empirical formula and molecular formula? Atomic masses of hydrogen, carbon and chlorine are 1.008,12.000 and 35.453 u, respectively
Solution:
Given: Percentage of H, C and Cl = 4.07% , 24.27% and 71.65% by mass respectively.
To find: Empirical formula and molecular formula
Calculation: Step I:
Check whether the sum of all the percentages is 100.
4.07 + 24.27 + 71.65 = 99.99 ≈ 100
Therefore, no need to consider presence of oxygen atom in the molecule.

Step II:
Conversion of mass percent to grams. Since we are having mass percent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07 g hydrogen 24.27 g carbon and 71.65 g chlorine are present.

Step III:
Convert into number of moles of each element. Divide the masses obtained above by respective atomic masses of various elements.
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}\) = 4.04 mol
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.000 \mathrm{~g}}\) = 2.0225 mol
Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}\) = 2.021 mol

Steps IV:
Divide the mole values obtained above by the smallest value among them.

Hence, the ratio of number of moles of 2 : 1 : 1 for H : C : Cl.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step V:
Write empirical formula by mentioning the numbers after writing the symbols of respective elements. CH₂Cl is thus, the empirical formula of the above compound.

Step VI:
Writing molecular formula
a. Determine empirical formula mass: Add the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass = 12.000 + 2 × 1.008 + 35.453 = 49.469 g mol^-1
b. Divide molar mass by empirical formula mass

c. Multiply empirical formula by r obtained above to get the molecular formula:
Molecular formula = r × empirical formula
∴ Molecular formula is 2 × CH₂Cl i.e. C₂H₄Cl₂.
Ans: The empirical formula of the compound is CH₂Cl and the molecular formula of the compound is C₂H₄Cl₂.
[Note: The question is modified to include the determination of molecular formula of the compound.]

Question 42.
A compound with molar mass 159 was found to contain 39.62% copper and 20.13% sulphur. Suggest molecular formula for the compound (Atomic masses: Cu = 63, S = 32 and O = 16).
Solution:
Given: Atomic mass of Cu = 63, S = 32, and O = 16
Percentage of copper and sulphur = 39.62% and 20.13% respectively.
To find: The molecular formula of the compound
Calculation: % copper + % sulphur = 39.62 + 20.13 = 59.75
This is less than 100 %. Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, % of oxygen = 100 – 59.75 = 40.25%

Hence, empirical formula is CuSO₄.
Empirical formula mass = 63 + 32 +16 × 4 = 159 g mol^-1
Hence,
Molar mass = Empirical formula mass
∴ Molecular formula = Empirical formula = CuSO₄
Ans: Molecular formula of the compound = CuSO₄

Question 43.
An inorganic compound contained 24.75% potassium and 34.75% manganese and some other common elements. Give the empirical formula of the compound. (K = 39 u, Mn = 54.9 u, O = 16 u)
Solution:
Given: Atomic mass of K = 39 u, Mn = 59 u, and O = 16 u.
Percentage of potassium and manganese = 24.75 % and 34.75% respectively
To find: The empirical formula of the given inorganic compound
Calculation: Percentage of potassium = 24.75%
Percentage of manganese = 34.75%
Total percentage = 59.50%
∴ Remaining must be that of oxygen
∴ Percentage of oxygen = 100 – 59.50 = 40.50%
Moles of K = \(\frac{\% \text { of } \mathrm{K}}{\text { Atomic mass of } \mathrm{K}}=\frac{24.75}{39}\) = 0.635 mol

Empirical formula = KMnO₄
Ans: The empirical formula of given inorganic compound is KMnO₄.

Question 44.
An organic compound contains 40.92% carbon by mass, 4.58% hydrogen and 54.50% oxygen. Determine the empirical formula of the compound.
Solution:
Given: Percentage mass of carbon = 40.92%; Percentage mass of hydrogen = 4.58%
Percentage mass of oxygen = 54.50%
To find: The empirical formula of compound
Calculation:

∴ Ratio = 1 : 1.34 : 1
Multiply by 3 to get whole number
∴ Ratio = 3 : 4.02 : 3 ≈ 3 : 4 : 3
∴ The empirical formula of the compound is C₃H₄O₃.
Ans: Empirical formula of the compound is C₃H₄O₃.

Question 45.
Define stoichiometric calculations.
Answer:
Calculations based on a balanced chemical equation are known as stoichiometric calculations.

Question 46.
Give reason: Balanced chemical equation is useful in solving problems based on chemical equations.
Answer:
Balanced chemical equation is symbolic representation of a chemical reaction. It provides the following information, which is useful in solving problems based on chemical equations:

• It indicates the number of moles of the reactants involved in a chemical reaction and the number of moles of the products formed.
• It indicates the relative masses of the reactants and products linked with a chemical change, and
• It indicates the relationship between the volume/s of the gaseous reactants and products, at STP.

Hence, balanced chemical equation is useful in solving problems based on chemical equations.

Question 47.
What are different types of stoichiometric problems? Write steps involved in solving stoichiometric problems.
Answer:
i. Generally, problems based on stoichiometry are of the following types:

• Problems based on mass-mass relationship
• Problems based on mass-volume relationship
• Problems based on volume-volume relationship.

ii. Steps involved in problems based on stoichiometric calculations:

• Write down the balanced chemical equation representing the chemical reaction.
• Write the number of moles and the relative masses or volumes of the reactants and products below the respective formulae.
• Relative masses or volumes should be calculated from the respective formula mass referring to the condition of STP.
• Apply the unitary method to calculate the unknown factors) as required by the problem.

Question 48.
Calculate the mass of carbon dioxide and water formed on complete combustion of 24 g of methane gas. (Atomic masses, C = 12 u, H = 1 u, O = 16 u)
Solution:
Mass of methane consumed in reaction = 24 g
Atomic mass: C = 12 u, H = 1 u, O = 16 u
To find: Mass of carbon dioxide and water formed
Calculation: The balanced chemical equation is,

Hence, 16 g of CH₄ on complete combustion will produce 44 g of CO₂.
∴ 24 g of CH₄ = \(\frac {24}{16}\) × 44 = 66 g of CO₂
Similarly, 16 g of CH₄ will produce 36 g of water.
∴ 24 g of CH₄ = \(\frac {24}{16}\) × 36 = 54 g of water
Ans: Mass of carbon dioxide and water formed respectively are 66 g and 54 g.

Question 49.
How much CaO will be produced by decomposition of 5 g CaCO₃?
Solution:
Given: Mass of CaCO₃ consumed in reaction = 5 g
To find: Mass of CaO produced
Calculation: Calcium carbonate decomposes according to the balanced equation,

So, 100 g of CaCO₃ produce 56 g of CaO.

Ans: Mass of CaO produced = 2.8 g

Question 50.
How many litres of oxygen at STP are required to burn completely 2.2 g of propane, C₃H₈ ?
Solution:
Given: Mass of propane used up in reaction = 2.2 g
To find: Volume of oxygen required at STP
Calculation:
The balanced chemical equation for the combustion of propane is,

(∵ 1 mol of ideal gas occupies 22.4 L of volume at STP)
Thus, 44 g of propane require 112 litres of oxygen at STP for complete combustion.
∴ 2.2 g of propane will require
\(\frac {112}{44}\) × 2.2 = 5.6 litres of O₂ at STP for complete combustion.
Ans: Volume of O₂ (at STP) required to bum 2.2 g propane = 5.6 litres

Question 51.
A piece of zinc weighing 0.635 g when treated with excess of dilute H₂SO₄ liberated 200 cm³ hydrogen at STP. Calculate the percentage purity of the zinc sample.
Solution:
Given: Mass of zinc = 0.635 g, volume of H₂ liberated = 200 cm³
To find: % purity of zinc sample
Calculation: The relevant balanced chemical equation is,
Zn + H₂SO₄ → ZnSO₄ + H₂
It indicates that 22.4 L of hydrogen at STP = 65 g of Zn.
(where, atomic mass of Zn = 65 u)
∴ 0.200 L of hydrogen at STP
= \(\frac{65 \mathrm{~g}}{22.4 \mathrm{~L}}\) × 200 L
= 0.5803 g of Zn
∴ Percentage purity of Zn = \(\frac{0.5803}{0.635}\) × 100
= 91.37 % (by using log tables)
Ans: Percentage purity of Zn sample = 91.37%

[Calculation using log table:
\(\frac{65 \times 0.200}{22.4}\)
= Antilog₁₀ [log₁₀ (65) + log₁₀ (0.200) – log₁₀ (22.4)]
= Antilog₁₀ [1.8129 + \(\overline{1} .3010\) – 1.3502]
= Antilog₁₀ [latex]\overline{1} .7637[/latex] = 0.5803
\(\frac{0.5803}{0.635} \times 100=\frac{58.03}{0.635}\)
= Antilog₁₀ [log₁₀ (58.03) – log₁₀ (0.635)]
= Antilog₁₀ [1.7636 – \(\overline{1} .8028\)]
= Antilog₁₀ [1.9608] = 91.37]

Question 52.
Explain with the help of a chemical reaction how limiting reagent works.
Answer:

• Consider the formation of nitrogen dioxide (NO₂) from nitric oxide (NO) and oxygen.
  2NO_(g) + O_2(g) → 2NO_2(g)
• Suppose initially, we take 8 moles of NO and 7 moles of O₂.
• To determine the limiting reagent, calculate the number of moles NO₂ produced from the given initial quantities of NO and O₂. The limiting reagent will yield the smaller amount of the product.
• Starting with 8 moles of NO, the number of NO₂ produced is,
  8 mol NO × \(\frac{2 \mathrm{~mol} \mathrm{NO}_{2}}{2 \mathrm{~mol} \mathrm{NO}}\) = 8 mol NO₂
• Starting with 7 moles of O₂, the number of moles NO₂ produced is,
  7 mol O₂ × \(\frac{2 \mathrm{~mol} \mathrm{NO}_{2}}{1 \mathrm{~mol} \mathrm{NO}}\) = 14 mol NO₂
• Since, 8 moles NO result in a smaller amount of NO₂, NO is the limiting reagent, and O₂ is the excess reagent, before reaction has started.

Question 53.
Urea [(NH₂)₂CO] is prepared by reacting ammonia with carbon dioxide.
2NH_3(g) + CO_2(g) → (NH₂)₂CO_(aq) + H₂O_(l)
In one process, 637.2 g of NH₃ are treated with 1142 g of CO₂.
i. Which of the two reactants is the limiting reagent?
ii. Calculate the mass of (NH₂)₂CO formed.
iii. How much excess reagent (in grams) is left at the end of the reaction?
Solution:
i. If 637.2 g of NH₃ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation.
Mass of NH₃ → Moles of NH₃ → Moles of (NH₂)₂CO

If 1142 g of CO₂ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation.
Mass of CO₂ → Moles of CO₂ → Moles of (NH₂)₂CO

Since NH₃ produces smaller amount of (NH₂)₂CO, the limiting reagent is NH₃.

iii. Starting with 18.71 moles of (NH₂)₂CO, we can determine the mass of CO₂ that reacted using the mole ratio from the balanced equation and the molar mass of CO₂ by the following relation.
Moles of (NH₂)₂CO → Moles of CO₂ → Grams of CO₂

The amount of CO₂ remaining = 1142 g – 823.4 g = 318.6 g ≈ 319 g CO₂ remaining
Ans: i. Limiting reagent = NH₃
ii. Mass of (NH₂)₂CO produced = 1124 g
iii. Mass of CO₂ remaining = 319 g

Question 54.
6 g of H₂ reacts with 32 g of O₂ to yield water. Which is the limiting reactant? Find the mass of water produced and the amount of excess reagent left.
Solution:
i. The reaction is: 2H_2(g) + O_2(g) → 2H₂O_(l)
If 6 g of H₂ react completely, calculate the number of moles of H₂O, that could be produced, by the following relation.
Mass of H₂ → Moles of H₂ → Moles of H₂O

If 32 g of O₂ react completely, calculate the number of moles of H₂O, that could be produced, by the following relation.
Mass of O₂ → Moles of O₂ → Moles of H₂O

Since O₂ produces smaller amount of H₂O, the limiting reagent is O₂.

iii. Starting with 2 moles of H₂O, we can determine the mass of H₂ that reacted using the mole ratio from the balanced equation and the molar mass of H₂ by the following relation.
Moles of H₂O → Moles of H₂ → Grams of H₂

The amount of H₂ remaining = 6g – 4g = 2g H₂ remaining
Ans: i. Limiting reagent = O₂
ii. Mass of H₂O produced = 36 g
iii. Mass of H₂ remaining = 2 g

Question 55.
Name different ways to express the concentration of a solution (or the amount of substance in a given volume of solution).
Answer:

• Mass percent or weight percent (w/w %)
• Mole fraction
• Molarity (M)
• Molality (m)

Question 56.
State the formula to obtain mass percent.
Answer:
Mass percent (w/w %) = \(\frac{\text { Mass of Solute }}{\text { Massof solution }} \times 100\)

Question 57.
Why is molality NOT affected by temperature?
Answer:

• Molality is the number of moles of solute present in 1 kg of solvent. Therefore, molality is mass dependent.
• Mass remains unaffected with temperature.

Hence, molality is not affected by temperature.

Question 58.
State TRUE or FALSE. Correct the statement if false.
i. A majority of reactions in the laboratory are carried out in in gaseous forms.
ii. Molarity is the most widely used to express the concentration of solution.
iii. Molality of a solution changes with temperature.
Answer:
i. False
A majority of reactions in the laboratory are carried out in solution forms.
ii. Tme
iii. False
Molality of a solution does not change with temperature.

Question 59.
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass percent of the solute.
Solution:
Given: Mass of substance = 2 g, mass of water = 18 g
To find: Mass % of solute

Ans: Mass percent of A = 10 % w/w

Question 60.
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Solution:
Given: Mass of solute (NaOH) = 4 g, volume of solution = 250 mL = 0.250 L
To find: Molarity of the solution

Ans: Molarity of the NaOH solution = 0.4 M

Question 61.
The density of 3 M solution of NaCl is 1.25 g mL^-1. Calculate molality of the solution.
Solution:
Given: Molarity of the solution = 3 M, density of the solution = 1.25 g mL^-1
To find: Molality of the solution
Formula:

Calculation: Molarity = 3 mol L^-1
∴ Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1 L solution = 1000 × 1.25 = 1250 g (∵ Density = 1.25 g mL^-1)
Mass of water in solution = 1250 – 175.5 = 1074.5 g = 1.0745 kg

Ans: Molality of the NaCl solution = 2.790 m

[Calculation using log table:
\(\frac{3}{1.0745}\)
= Antilog₁₀ [log₁₀ (3) – log₁₀ (1.0745)]
= Antilog₁₀ [0.4771 – 0.0315]
= Antilog₁₀ [0.4456] = 2.790]

Question 62.
Calculate the molarity of 1.8 g HNO₃ dissolved in 250 mL aqueous solution.
Solution:
Mass of HNO₃ = 1.8 g,
volume of solution = 250 mL = 0.250 L
To find: Molarity
Formulae:

Calculation: Molar mass of HNO₃ = 63 g mol^-1
Using formula (i),

Ans: Molarity of the HNO₃ solution = 0.114 M

Question 63.
What volume in mL of a 0.1 M H₂SO₄ solution will contain 0.5 moles of H₂SO₄?
Solution:
Given: Molarity of H₂SO₄ solution = 0.1 M
Moles of H₂SO₄ = 0.5 mol
To find: Volume of H₂SO₄ solution

= 5 L
= 5000 L
Ans: Volume of a 0.1 M H₂SO₄ solution = 5000 mL

Question 64.
A mixture has 18 g water and 414 g ethanol. What are the mole fractions of water and ethanol?
Solution:
Given: Mass of water = 18 g, mass of ethanol = 414 g
To find: Mole fractions of water and ethanol
Formulae:


Ans: Mole fraction of water = 0.1
Mole fraction of ethanol = 0.9

Question 65.
Explain in brief: Use of graphs in analytical chemistry.
Answer:
i. Analytical chemistry often involves deducing some relation between two or more properties of matter under study.
ii. For example, the relation between temperature and volume of a given amount of gas.
iii. A set of experimentally measured values of volume and temperature of a definite mass of a gas upon plotting on a graph paper appears as in the figure below:

iv. When the points are directly connected, a zig zag pattern results as shown below:

From the above pattern, no meaningful result can be deduced.
v. A smooth curve (or average curve) passing through these points can be drawn as shown below. This straight line is consistent with the V ∝ T .

vi. While fitting the points into a smooth curve, all the plotted points should be evenly distributed. This can be verified mathematically, by drawing a perpendicular from each point to the curve. The perpendicular represents deviation of each point from the curve. Take sum of all the perpendiculars on side of the line and sum of all the perpendiculars on another side of the line separately. If the two sums are equal (or nearly equal), the curve drawn shows the experimental points in the best possible representation

Question 66.
1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO₄ was obtained as dry precipitate. Calculate the percentage purity of the sample.
Solution:

The chemical equation representing the reaction is:

To calculate the mass of Na₂SO₄ from which 1.74 g of BaSO₄ is obtained:
233 g of BaSO₄ is produced from 142 g of Na₂SO₄.
∴ Mass of Na₂SO₄ from which 1.74 g of BaSO₄ would be obtained = \(\frac {142}{233}\) × 1.74 = 1.06 g
∴ The mass of pure Na₂SO₄ present in 1.5 g of impure sample = 1.06 g
To calculate the percentage purity of the impure sample:
1.5 g of impure sample contains 1.06 g of pure Na₂SO₄
∴ 100 g of the impure sample will contain = \(\frac{1.06}{1.5}\) × 100 = 70.67 g of pure Na₂SO₄
Ans: Percentage purity of the sample is 70.67 %.

Question 67.
Calculate the amount of lime Ca(OH)₂, required to remove hardness of 50,000 L of well water which has been found to contain 1.62 g of calcium bicarbonate per 10 L .
Solution:
Calculation of total Ca(HCO₃)₂ present:
10 L of water contains 1.62 g of Ca(HCO₃)
∴ 50,000 L of water will contain \(\frac{1.62}{10}\) × 50,000 = 8100 g of Ca(HCO₃)
Calculation of lime required:
The balanced equation for the reaction:

∴ 162 g of Ca(HCO₃) requires 74 g of lime.
Mass of lime required by 8100 g of Ca(HCO₃) = \(\frac {74}{162}\) × 8100 g = 3700 g = 3.7 kg
Ans: The amount of lime required to remove hardness of 50,000 L of well water, with 1.62 g of calcium bicarbonate per 10 L, is 3.7 kg.

Multiple Choice Questions

1. The number of significant figures in 0.0110 is …………….
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

2. For the following measurements in which the true value is 4.0 g, find the CORRECT statement.

Student	Readings (g)
A	4.01	3.99
B	4.05	3.95

(A) Results of both the students are neither accurate nor precise.
(B) Results of student A are both precise and accurate.
(C) Results of student A are neither precise nor accurate.
(D) Results of student B are both precise and accurate.
Answer:
(B) Results of student A are both precise and accurate.

3. 18.238 is rounded off to four significant figures as ………….
(A) 18.20
(B) 18.23
(C) 18.2360
(D) 18.24
Answer:
(D) 18.24

4. The % of H₂O in Fe(CNS)₃.3H₂O is ……………..
(A) 34
(B) 11
(C) 19
(D) 46
Answer:
(C) 19

5. The molecular mass of an organic compound is 78 g mol^-1. Its empirical formula is CH. The molecular formula is ………….
(A) C₂H₄
(B) C₂H₂
(C) C₆H₆
(D) C₄H₄
Answer:
(C) C₆H₆

6. The percentage of oxygen in NaOH is ……………
(A) 40%
(B) 60%
(C) 8%
(D) 10%
Answer:
(A) 40%

7. 1.2 g of Mg (At. mass 24) will produce MgO equal to …………….
(A) 0.05 mol
(B) 0.03 mol
(C) 0.01 mol
(D) 0.02 mol
Answer:
(A) 0.05 mol

8. …………. reagent is the reactant that reacts completely but limits further progress of the reaction.
(A) Oxidizing
(B) Reducing
(C) Limiting
(D) Excess
Answer:
(C) Limiting

9. If a solution is made up of 1 mol ethanol and 9 mol water, then mole fraction of water in the solution is ……………
(A) 0.1
(B) 0.5
(C) 0.9
(D) 1.0
Answer:
(C) 0.9

10. 0.9 glucose (C₆H₁₂O₆) is present in 1 L of solution. Find molarity.
(A) 5 M
(B) 50 M
(C) 0.005 M
(D) 0.5 M
Answer:
(C) 0.005 M

11. Molality of a solution is the ……………
(A) number of moles of solute present in 1 kg of solvent
(B) number of moles of solute present in 1 L of solution
(C) mass of solute present in 1 kg of solvent
(D) number of moles of solute present in 1 kg of solution
Answer:
(A) number of moles of solute present in 1 kg of solvent

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 7 Depreciation Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 7 Depreciation

1. Answer in One Sentence only.

Question 1.
What is the Reducing Balance Method of charging depreciation?
Answer:
A method of charging depreciation in which depreciation is charged on fixed assets at a fixed percentage on its opening balance every year is called the reducing balance method.

Question 2.
Under which method of depreciation the amount of depreciation remains constant every year?
Answer:
Under the Fixed Instalment Method of depreciation, an amount of depreciation remains constant every year.

Question 3.
Under which method of depreciation does the number of depreciation change every year?
Answer:
Under the Reducing Balance Method of depreciation amount of depreciation changes every year.

Question 4.
Which method of depreciation would you suggest for depreciating a five years lease?
Answer:
For depreciating a five years lease fixed installment Method of depreciation is suggested.

Question 5.
What is meant by the cost of assets?
Answer:
The sum total of the purchase price of a fixed asset and its installation charges are called the cost of the asset.

2. Write the word/term/phrase which can substitute each of the following statements:

Question 1.
The method of charging depreciation under which depreciation is calculated on the original cost of an asset.
Answer:
Fixed Instalment Method

Question 2.
The method of charging depreciation under which depreciation is calculated on the balance amount.
Answer:
Reducing Balance Method

Question 3.
The Latin word for reduction or decline in the value of a fixed asset due to its use.
Answer:
Depretium

Question 4.
An amount to which the balance in the depreciation account is transferred.
Answer:
Profit and Loss A/c

Question 5.
Transport charges, coolie charges, charges for electrification, etc. incurred for the erection of machinery.
Answer:
Installation Charges

3. Select the most appropriate answers from the alternatives given below and rewrite the sentences.

Question 1.
Wages paid for installation of machinery is debited to ____________ account.
(a) Profit and Loss
(b) Trading
(c) Wages
(d) Machinery
Answer:
(d) Machinery

Question 2.
Depreciation arises because of ____________
(a) wear and tear
(b) inflation
(c) loss in business
(d) profit in business
Answer:
(a) wear and tear

Question 3.
The profit on sale of an asset is debited to ____________ A/c.
(a) Profit and Loss
(b) Reserve
(c) Asset
(d) Balance sheet
Answer:
(c) Asset

Question 4.
By the amount of depreciation the value of asset ____________
(a) decreases
(b) increases
(c) becomes zero
(d) remains constant
Answer:
(a) decreases

Question 5.
In fixed instalment system the amount of depreciation is ____________ every year.
(a) constant
(b) fluctuating
(c) increased
(d) decreased
Answer:
(a) constant

Question 6.
Under ____________ method depreciation is calculated on written down value.
(a) Fixed Instalment
(b) Reducing Balance
(c) Revaluation
(d) Depletion
Answer:
(b) Reducing Balance

Question 7.
Under the ____________ system of depreciation, the amount of depreciation does not change from year to year.
(a) Fixed Instalment
(b) Reducing Balance
(c) Depletion
(d) Machine Hour Rate
Answer:
(a) Fixed Instalment

Question 8.
Depreciation = \(\frac{Cost of Asset (-) ………….}{Estimated life of Asset}\)
(a) Purchase price
(b) Scrap value
(c) Installation charges
(d) months
Answer:
(b) Scrap value

4. State whether the following statements are True or False with reasons.

Question 1.
There is no need to provide depreciation if the asset is maintained with care.
Answer:
This statement is False.
There is no relation between good maintenance of assets and depreciation working life of fixed assets decreases with the passage of time and the introduction of new technology. So even assets are maintained with care depreciation is provided.

Question 2.
Under the Reducing Balance, method depreciation is charged on the original cost.
Answer:
This statement is False.
Underwritten down value method depreciation is calculated at a certain fixed rate of percentage every on the balance of the asset which is brought forward from the previous year.

5. Complete the following sentence.

Question 1.
____________ is the major cause for Depreciation.
Answer:
Normal and Natural wear of Tear

Question 2.
Depreciation is ____________ to the business.
Answer:
Loss

Question 3.
Depreciation is necessary to make provision for ____________ of old assets.
Answer:
replacement

Question 4.
Depreciation enables the business to compute and pay the correct amount of ____________ to the Government.
Answer:
Tax

Question 5.
____________ cost concept is use for depreciation of Assets.
Answer:
Historical

Question 6.
Fixed Installment Method of depreciation is also known as ____________ cost method.
Answer:
Original

Question 7.
____________ method of depreciation is recognised by Tax/Law.
Answer:
Written Down

Question 8.
____________ account is credited when incidental expenses paid on acquiring Assets.
Answer:
Cash/Bank

Question 9.
Balance of depreciation A/c is transferred to ____________ A/c at the end of financial year.
Answer:
profit and loss

Question 10.
Depreciation is ____________ expenses.
Answer:
non cash

6. Do you agree or disagree with the following statements.

Question 1.
Depreciation is charged on current Assets only.
Answer:
Disagree

Question 2.
Depreciation is not charged on Intangible assets.
Answer:
Disagree

Question 3.
No depreciation is charged on Land.
Answer:
Agree

Question 4.
Written Down value of the asset is calculated by adding depreciation for a period of use of assets.
Answer:
Disagree

Question 5.
No depreciation is charged on assets purchased on credit.
Answer:
Disagree

7. Correct the following statement and rewrite the statement.

Question 1.
Depreciation is cash expenses.
Answer:
Depreciation is a non-cash expense.

Question 2.
The depreciation account is a Real account.
Answer:
The depreciation account is a Nominal account.

Question 3.
Wages paid on the installation of Machinery is Debited to wages A/c.
Answer:
Wages paid on the installation of Machinery is Debited to Machinery A/c.

Question 4.
Depreciation is charged only when the business is making a loss.
Answer:
Depreciation is charged whether a business is making a profit or loss.

Question 5.
Depreciation increases the value of an asset.
Answer:
Depreciation reduces the value of the Asset.

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 13 Organisms and Populations Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 13 Organisms and Populations

Multiple choice questions

Question 1.
All ecosystems on the earth together form
(a) biosphere
(b) biome
(c) living world
(d) environment
Answer:
(a) biosphere

Question 2.
What is the mean annual temperature in the region of Arctic and Alpine tundra ?
(a) About -10 to 2 °C
(b) About -2 to 2 °C
(c) About 0 to 5 °C
(d) About 5 to 10 °C
Answer:
(a) About -10 to 2 °C

Question 3.
Which out of the following are the major biomes in India ?
(I) Desert (II) Grassland (III) Tropical rain forest (IV) Temperate forest (V) Coniferous forest (VI) Deciduous forest (VII) Sea coast (VIII) Tundra
(a) (II) (IV) (VI) (VII)
(b) (I) (II) (V) (VIII)
(c) (I) (III) (VI) (VII)
(d) (II) (III) (VI) (VIII)
Answer:
(c) (I) (III) (VI) (VII)

Question 4.
Important key elements that bring about variations in the different habitats are
(a) temperature, water, light and soil
(b) salinity, pollutants, water
(c) modern developmental parameters in the region
(d) scientific progress and technological innovations
Answer:
(a) temperature, water, light and soil

Question 5.
Which out of the following is the most ecologically relevant factor?
(a) Water
(b) Temperature
(c) Salinity
(d) Wind speed
Answer:
(b) Temperature

Question 6.
Animals that can tolerate a narrow range of temperature are
(a) stenothermal
(b) eurythermal
(c) poikilothermic
(d) homeothermic
Answer:
(a) stenothermal

Question 7.
Animals that can tolerate a narrow range of salinity are
(a) euryhaline
(b) anadromous
(c) catadromous
(d) stenohaline
Answer:
(d) stenohaline

Question 8.
What is the source of energy in the depths of more than 500 m in the oceans?
(a) Sunlight
(b) Wind energy
(c) Dead and decaying matter
(d) Phytoplankton
Answer:
(c) Dead and decaying matter

Question 9.
Which factor does not determine percolation and water holding capacity of the soil?
(a) Soil composition
(b) Grain size
(c) Aggregation of soil particles
(d) Vegetation on that soil
Answer:
(d) Vegetation on that soil

Question 10.
Which factors of soil does not determine the vegetation in any area?
(a) pH
(b) mineral composition
(c) topography
(d) types of microorganisms
Answer:
(d) types of microorganisms

Question 11.
Find the odd one out
(a) Dormancy
(b) Hibernation
(c) Aestivation
(d) Migration
Answer:
(d) Migration

Question 12.
Which of the following ability is present in the desert animals?
(a) Ability to concentrate urine.
(b) Ability to remain inside the shelters and escape need of water.
(c) Ability to derive water from all the fruits.
(d) Ability to absorb water from air.
Answer:
(a) Ability to concentrate urine.

Question 13.
Which of the statement does not describe the adaptation of the desert plants ?
(a) Desert plants have a thick cuticle on their leaf surfaces.
(b) Desert plants have their stomata arranged in sunken pits.
(c) Desert plants have a special photosynthetic pathway called CAM.
(d) Desert plants have soft stems and large leaves.
Answer:
(d) Desert plants have soft stems and large leaves.

Question 14.
What is the use of CAM type of photosynthetic pathway for the desert plants ?
(a) It enables their stomata to remain closed during day time.
(b) It requires less sunlight for the photosynthesis.
(c) It requires less amount of chlorophyll during photosynthesis.
(d) The water absorbed from the soil by the plants during CAM photosynthetic pathway is less.
Answer:
(a) It enables their stomata to remain closed during day time.

Question 15.
In which plant photosynthetic function is taken over by the flattened stems?
(a) Nephrolepis
(b) Cycas
(c) Opuntia
(d) Zea mays
Answer:
(c) Opuntia

Question 16.
What is Allen’s rule?
(a) Mammals from colder climates generally have shorter ears and limbs to minimise heat loss.
(b) Mammals have constant temperature of the body in spite of their varied habitats.
(c) Mammals can be oviparous at times.
(d) The ecosystem of cold climate regions is equally occupied with mammals, birds and reptiles.
Answer:
(a) Mammals from colder climates generally have shorter ears and limbs to minimise heat loss.

Question 17.
Sunken stomata is the characteristic feature of
(a) hydrophyte
(b) mesophyte
(c) xerophyte
(d) halophyte
Answer:
(c) xerophyte

Question 18.
Which of the following pairs is correctly matched ?
(a) Uricotelism-aquatic habitat
(b) Parasitism-intra-specific relationship
(c) Excessive perspiration-xeric adaptation
(d) Stream-lined body-aquatic adaptation
Answer:
(d) Stream-lined body-aquatic adaptation

Question 19.
World population day is observed on
(a) 11th July
(b) 16th September
(c) 23rd October
(d) 1st December
Answer:
(a) 11th July

Question 20.
World Environment day is celebrated on
(a) 22nd April
(b) 5th June
(c) 1st October
(d) 16th November
Answer:
(b) 5th June

Question 21.
World Earth day is observed on
(a) 22nd April
(b) 5th June
(c) 21st October
(d) 26th November
Answer:
(a) 22nd April

Question 22.
World ozone day is celebrated on
(a) 5th June
(b) 16th September
(c) 1st October
(d) 23rd October
Answer:
(b) 16th September

Question 23.
A group of individuals belonging to the same species within an ecosystem is called a
(a) community
(b) habitat
(c) population
(d) specific group
Answer:
(c) population

Question 24.
The populations of different species that live and interact together in the ecosystem are called
(a) community
(b) habitat
(c) population
(d) interspecies
Answer:
(a) community

Question 25.
If do not occur in an ecosystem, the survival of organisms may not take place and functioning of ecosystem is lost.
(a) interactions
(b) struggle
(c) reproduction
(d) none of the above
Answer:
(a) interactions

Question 26.
If in a pond there were 200 lotus plants last year and through reproduction 20 new plants are added, taking the current population to 220, what is the natality rate of the lotus for that year ?
(a) 0.2
(b) 0.4
(c) 0.1
(d) 1.0
Answer:
(c) 0.1

Question 27.
During laboratory experiments, 30 fishes died from an aquarium tank having 150 fishes during one month. What is the rate of mortality of fishes per month ?
(a) 0.2
(b) 0.3
(c) 0.4
(d) 0.5
Answer:
(a) 0.2

Question 28.
Which of the following is correct statement?
(a) Natality can never be controlled in any population.
(b) If mortality is more than natality, the density of population declines.
(c) Natality and mortality are always same for every population.
(d) If natality is more than mortality the population size declines.
Answer:
(b) If mortality is more than natality, the density of population declines.

Question 29.
What is the apt definition of density?
(a) The capacity of a population to sustain.
(b) The total number of genes in a population.
(c) The total number of individuals in a population per unit area.
(d) The total number of births taking place.
Answer:
(c) The total number of individuals in a population per unit area.

Question 30.
On which of the following factors is growth rate of a population dependent ?
(a) Density and natality
(b) Natality and age structure
(c) Mortality and density
(d) Natality and mortality
Answer:
(a) Density and natality

Question 31.
When pre-reproductive age group is large in a population, what will be the growth rate of that population?
(a) Steady
(b) Rapid
(c) Declining
(d) None of these
Answer:
(b) Rapid

Question 32.
When pre-reproductive and post-reproductive age group is same in structure, the population is
(a) declining
(b) increasing
(c) steady
(d) disappearing
Answer:
(c) steady

Question 33.
In which type of interactions, interacting species do not live closely together ?
(a) Competition
(b) Parasitism
(c) Commensalism
(d) Predation
Answer:
(d) Predation

Question 34.
Name the interaction in which one species is harmed but the other remains unaffected,
(a) Commensalism
(b) Parasitism
(c) Amensalism
(d) Competition
Answer:
(c) Amensalism

Question 35.
Choose the correct statement from the following
(a) Carnivores are the only predators in any ecosystem
(b) Herbivores are in a broad ecological context not very different from predators.
(c) Sparrow eating grain is not a predator.
(d) Predation and parasitism are one and the same.
Answer:
(b) Herbivores are in a broad ecological context not very different from predators.

Question 36.
What is the meaning of symbiosis?
(a) Living together
(b) Breeding together
(c) Fighting with each other
(d) Feeding together
Answer:
(a) Living together

Question 37.
Gause’s ‘Competitive Exclusion Principle’ can work only when
(a) the resources are limited
(b) the resources are abundant
(c) the inferior species and superior species do not need same resources
(d) only when there is interspecific competition
Answer:
(a) the resources are limited

Question 38.
The word commensalism means
(a) on the either side of the table
(b) sharing the table
(c) eating at the table of other
(d) on the top of the table
Answer:
(b) sharing the table

Question 39.
Which of the following is not a classical example of commensalism?
(a) An orchid growing as an epiphyte on a mango branch.
(b) Barnacles growing on the back of a whale.
(c) The cattle egret and grazing cattle.
(d) Lichen seen on the tree.
Answer:
(d) Lichen seen on the tree

Question 40.
Calotropis plant is poisonous for herbivores as it is rich in
(a) opium
(b) cardiac glycosides
(c) quinine
(d) strychnine
Answer:
(b) cardiac glycosides

Question 41.
Term ecology was first used by
(a) Grinnell
(b) Reiter
(c) Haeckel
(d) Darwin
Answer:
(b) Reiter

Question 42.
Who introduced subject ecology to the world.
(a) Reiter
(b) Grinnell
(c) Darwin
(d) E. Haeckel
Answer:
(d) E. Haeckel

Question 43.
The term niche was first time used by
(a) Grinnell
(b) Haeckel
(c) Mendel
(d) Darwin
Answer:
(a) Grinnell

Match the columns

Question 1.

Column A	Column B
(1) Organism	(a) Large unit with specific climatic zone
(2) Population	(b) Different species in particular area
(3) Community	(c) Same species in a geographical area
(4) Biome	(d) Basic unit of ecological hierarchy

Answer:

Column A	Column B
(1) Organism	(d) Basic unit of ecological hierarchy
(2) Population	(c) Same species in a geographical area
(3) Community	(b) Different species in particular area
(4) Biome	(a) Large unit with specific climatic zone

Question 2.

Water body A	Salinity values B
(1) Streams	(a) 340 ppt
(2) Indian Ocean	(b) 5 ppt
(3) Hyper saline lagoon	(c) 30-35 ppt
(4) Dead sea	(d) 100 ppt

Answer:

Water body A	Salinity values B
(1) Streams	(b) 5 ppt
(2) Indian Ocean	(c) 30-35 ppt
(3) Hyper saline lagoon	(d) 100 ppt
(4) Dead sea	(a) 340 ppt

Classify the following to form Column B as per the category given in Column A

Question 1.
Emigration, Immigration, More mortality, Unlimited resources, More natality, Continuous reproduction.

Column A	Column B
(1) Decrease in population density	————–
(2) Increase in population density	—————
(3) Exponential growth	————–

Answer:

Column A	Column B
(1) Decrease in population density	Emigration, More mortality
(2) Increase in population density	Immigration, More natality
(3) Exponential growth	Unlimited resources, Continuous reproduction.

Question 2.
Hungary, United States, Denmark, Italy, Australia, Kenya, Nigeria, Germany

Pattern of population growth	Name of the countries
(1) Rapid growth	————–
(2) Slow growth	—————
(3) Zero growth	————–
(4) Negative growth	————–

Answer:

Pattern of population growth	Name of the countries
(1) Rapid growth	Kenya, Nigeria
(2) Slow growth	United States, Australia
(3) Zero growth	Denmark, Italy
(4) Negative growth	Hungary, Germany

Question 3.
Epiphytic orchid and mango tree, Ascaris and human, Egret and cattle, Lichen having alga and fungus, Bumblebee and flowering plant, Plasmodium vivax and man.

Column A	Column B
(1) Mutualism	————–
(2) Commensalism	—————
(3) Parasitism	————–

Answer:

Column A	Column B
(1) Mutualism	Lichen having alga and fungus, Bumblebee and flowering plant.
(2) Commensalism	Epiphytic orchid and mango tree, Egret and cattle.
(3) Parasitism	Ascaris and human, Plasmodium vivax and man.

Very short answer questions

Question 1.
What are the key abiotic factors that influence all habitats?
Answer:
Key abiotic factors that influence all habitats are ambient temperature, availability of water, light and type of soil.

Question 2.
What is homeostasis?
Answer:
Homeostasis is the state of steady interned, physical and chemical conditions maintained by living systems.

Question 3.
What is meant by eurythermal?
Answer:
Eurythermal are those organisms that can tolerate and thrive in a wide range of temperatures.

Question 4.
What is meant by stenothermal?
Answer:
Stenothermal are those organisms which are restricted to only narrow range of temperatures.

Question 5.
What is euryhaline?
Answer:
Organisms which can tolerate wide range of salinities are called euryhaline.

Question 6.
What is stenohaline?
Answer:
Organisms which are restricted only to a narrow range of salinity are called stenohaline.

Question 7.
What are the various characteristics of soil?
Answer:
Characteristics of the soil are soil composition, grain size, the percolation and water holding capacity, pH, mineral composition of the soil.

Question 8.
What decides the vegetation of an area?
Answer:
The soil characteristics along with pH, mineral composition and topography, and climatic factors determine the vegetation of an area.

Question 9.
What is Population ecology?
Answer:
Population ecology is an important area of ecology because it links ecology to population dynamics, genetics and evolution.

Question 10.
In which interaction of species, both the species are at a loss?
Answer:
Competition is the type of interaction where . both the species are at a loss.

Question 11.
What is parasitism?
Answer:
Parasitism is the type of interaction between two species in which parasitic species is benefited and the host species is harmed.

Question 12.
What are ectoparasites?
Answer:
Parasites that feed on the external surface of the host organism are called ectoparasites.

Question 13.
Name blood sucking ectoparasites.
Answer:
Lice, mosquito feeding on human blood and ticks parasitic on dogs.

Question 14.
Name the malarial parasite and its vector.
Answer:
The malarial parasite is Plasmodium vivax which needs a vector anopheles mosquito.

Question 15.
Name the secondary metabolites that act as defensive substances against grazers and browsers.
Answer:
Nicotine, caffeine, quinine, strychnine, opium, etc. are secondary metabolites which act as defensive substances produced by plants against grazers and browsers.

Question 16.
Name the ectoparasites which infest the marine fish.
Answer:
Copepods are the ectoparasites which infest the marine fish.

Question 17.
What do you mean by ‘evolutionary arms race’? In which kind of interactions is it observed?
Answer:
In order to save their lives, prey species : have evolved various defence mechanisms. This are called ‘evolutionary arms race’. These defence mechanisms are seen in prey-predator interactions.

Give Definitions of the following

Question 1.
Habitat
Answer:
The physical space of an organism with the other living or non-living factors is called its habitat.

Question 2.
Microhabitat
Answer:
The immediate surrounding of an organism is called microhabitat.

Question 3.
Niche
Answer:
Niche is defined as the processes about how that organism is linked with its physical and biological environment.

Question 4.
Fundamental niche
Answer:
Fundamental niche is the niche in the absence of all competitors, this is highly improbable in nature.

Question 5.
Realized niche
Realized niche is more realistic approach, in the presence of competition for the resources available in the habitat.

Question 6.
Competition
Answer:
Competition is defined as a process in which the fitness of one species is significantly lowered in the presence of another species.

Give one or two examples of the following

Question 1.
Intraspecific competition
Answer:
Two dogs fighting for same food. Two tomcats fighting for their territory.

Question 2.
Commensalism
Answer:
An orchid growing as an epiphyte on a branch of mango tree, is benefited due to support offered by mango tree but the mango tree does not get any benefit.

Question 3.
Ectoparasites
Answer:
Mosquito, louse sucks blood from human host.

Question 4.
Endoparasites
Answer:
Round worm. Ascaris and tape worm are endoparasites in the intestine of humans.

Name the type of association

Question 1.
Cattle egret birds with buffalo.
Answer:
Commensalism

Question 2.
Tiger and the deer.
Answer:
Predator and prey relationship

Question 3.
Visiting flamingos and fishes in the estuarine water.
Answer:
Interspecific competition.

Distinguish between the following

Question 1.
Natality and Mortality.
Answer:

Natality	Mortality
1. Birth rate of any population is called its natality.	1. Death rate of any population is called its mortality.
2. Rise in natality increase the population density.	2. Rise in mortality decrease the population density.
3. Decline in natality decrease the population.	3. Decline in mortality increase the population.
4. Natality is the positive factor for population growth.	4. Mortality is the negative factor for population growth.
5. Absolute natality will always be more than realized natality.	5. Absolute mortality will always be less them realized mortality.

Question 2.
Eurythermal and stenothermal.
Answer:

Eurythermal	Stenothermal
1. Animals which can tolerate wide range of temperatures are called eurythermal.	1. Animals which can tolerate only narrow range of temperature fluctuations are called stenothermal.
2. Eurythermal animals show reduced temperature sensitivity.	2. Stenothermal animals show high temperature sensitivity.
3. Body functions of eurythermal animals can occur at wide range of temperature range. E.g. Goat, man, cat, dog, tiger, cow, sheep, monkey, crab, etc.	3. Body functions of stenothermal animals can occur at only narrow range of temperature range. E.g. Insects, fishes, reptiles, snakes, etc.

Give scientific reasons

Question 1.
Temperature is said to be the most ecologically relevant environmental factor.
Answer:

1. Temperature fluctuations on the earth are quite marked.
2. The distribution of plants and animals on the earth depends upon temperature range.
3. For the organisms ambient temperature affects their enzyme kinetics of the cell.
4. Entire metabolism, activity and other physiology of the organism is dependent on temperature. Therefore, it is said to be the most ecologically relevant environmental factor.

Question 2.
Adaptation is an important attribute of the organism.
Answer:

1. Organisms adapt to their surrounding environment by showing physiological, behavioural or morphological changes which are called adaptations.
2. Due to adaptations, organisms can survive and reproduce in its environment. Therefore, adaptation is said to be am important attribute of the organisms.

Question 3.
Both host and the parasite tend to co¬evolve, against each other.
Answer:

1. Host and parasitic relationship is most specific. This means that for a particular parasite there is specific host.
2. Many parasites have evolved to be host-specific as they can parasitize only a single species of host. Therefore, during evolution, they both co-evolve together, against each other.

Question 4.
Cuscuta plant does not have chlorophyll.
Answer:

1. Cuscuta is a parasitic plant. It is commonly found growing on hedge plant.
2. It derives its nutrition directly from the host plant on which it thrives by parasitizing it.
3. Since it does not prepare its own food by photosynthesis, it loses chlorophyll from the leaves.

Question 5.
Predators in nature are called prudent.
Answer:

1. Predators control the prey population but if a predator over exploits its prey, then the prey might become extinct.
2. If prey species is not available, the predator will also starve and become extinct. Predators, therefore, do not kill the prey unnecessarily. They act as prudent.

Write short notes

Question 1.
Temperature fluctuations on the earth.
Answer:

1. The temperatures vary from subzero levels in polar areas and high altitudes, to about 50 °C in tropical deserts during summer.
2. There are also seasonal changes in the temperature.
3. Temperature also shows progressive decrease from the equator towards the poles and from plains to the mountain tops.
4. Some unique habitats such as hot springs may show very high temperatures of about 80 to 100 °C.
5. In deep-sea hydrothermal vents average temperatures may rise up to 400 °C.

Question 2.
Adaptations of mammals in colder regions.
Answer:

1. Mammals inhabiting colder regions have shorter snout, ears, tail and limbs to minimize the loss of body heat. This is called Allen’s Rule.
2. Aquatic mammals such as whales and seals living in the polar seas, have a thick layer of fat which is called a blubber below their skin.
3. Blubber acts as an insulator and thus helps to keep the body warm by reducing loss of body heat.
4. Some animals like polar bears undergo hibernation and thus tide over the stressful winters.

Question 3.
Natality.
Answer:

1. Natality is the birth rate of a population. Due to increased natality the population density rises.
2. Natality is a crude birth rate or specific birth rate.
3. Crude birth rate : Number of births per 1000 population/year gives crude birth rate. Crude birth rate is helpful in calculating population size.
4. Specific birth rate : Crude birth rate is relative to a specific criterion such as age. E.g. If in a pond, there were 200 carp fish and their population rises to 800. Then, taking the current population to 1000, the birth rate becomes 800/200 = 4 offspring per carp per year. This is specific birth rate.
5. Absolute Natality : The number of births under ideal conditions when there is no competition and the resources such as food and water are abundant, then it give absolute natality.
6. Realized Natality : The number of births under different environmental pressures give realized natality. Absolute natality will be always more than realized natality.

Question 4.
Mortality.
Answer:

1. Mortality is the death rate of a population. It gives a measure of the number of deaths in a particular population, in proportion to the size of that population, per unit of time.
2. Mortality rate is typically expressed in deaths per 1,000 individuals per year.
   A mortality rate of 9.5 (out of 1,000) in a population of 1,000 would mean 9.5 deaths per year in that entire population or 0.95% out of the total.
3. Absolute Mortality : The number of deaths under ideal conditions when there is no competition, and all the resources such as food and water are abundant, then it gives absolute mortality.
4. Realized Mortality : The number of deaths under environmental pressures come into play gives realized mortality.
5. It must be remembered that absolute mortality will always be less than realized mortality.

Question 5.
Populations Interactions
Answer:

1. In nature, every species requires interactions with at least one other species for its food.
2. Even autotrophic plant species needs soil microbes to break down the organic matter in soil and return the inorganic nutrients for absorption.
3. The plants need animal agents . for pollination.
4. Animals, plants and microbes cannot live in isolation but interact in various ways to form a biological community.
5. Such interactions can be interspecific (within two different species) or intraspecific (within the same species).
6. Interspecific interactions are of broad four types viz, neutralism, negative or harmful, positive or beneficial, and both positive and negative interactions.
7. Interacting species live closely together in interactions such as predation, parasitism and commensalism.
8. Neutralism interaction have no significant effect on either species. Negative interactions are of competition or amensalism type. Positive interactions occur in the form of mutualism, protocooperation and commensalism. Parasitism and predation are both positive and negative type of interaction.

Question 6.
Mutualism
Answer:

1. Mutualism is an obligatory and interdependent interaction. It is an association of two species in which both of them are benefited.
2. The classic example of mutualism is lichens. Lichen is an intimate, mutualistic relationship between a fungus and photosynthetic algae or cyanobacteria.
3. Most of the plant and animal interactions are of mutualistic type.
4. For pollination and seed dispersal, plants depend on the animals.
5. Animals in turn feed on pollen and nectar during pollination. During seed dispersal juicy and nutritious fruits are used by the animals.
6. In animal-animal interactions also mutualism is seen in many instances.

Question 7.
Brood parasitism
Answer:

1. Brood parasitism is a type of parasitic behaviour shown by Asian Koel. Koel lays its eggs in the nest of crow.
2. Crow acts as a host bird and incubate the eggs of koel.
3. The eggs of koel show resemblance to the host’s egg in size and colour. This reduces the chances of the crow detecting koel’s eggs and ejecting them from the nest.
4. Eggs of koel hatch before the host’s eggs and hence parasitic bird is in advantage.

Short answer questions

Question 1.
What are the three main types of niches?
Answer:

1. Spatial or habitat niche : Spatial or habitat niche means the physical space occupied by the organisms.
2. Trophic niche : This kind of niche is based on the trophic level of an organism in a food chain.
3. Multidimensional or hypervolume niche : In multidimensional niche, number of abiotic and biotic environmental factors are considered. The resulting space by the niche is called hypervolume. Therefore it is also called hypervolume niche.
4. It shows the position of an organism in the environmental gradient.

Question 2.
Why do animals need to maintain homeostasis?
Answer:

1. Homeostasis keeps the body in equilibrium.
2. All the internal functions are maintained due to homeostasis.
3. Survival, growth and reproduction can be achieved due to this steady state.
4. The external environment changes constantly but by homeostasis, organisms can cope up with this change.
5. Thus homeostasis is a way of adaptation for survival.

Question 3.
How is sunlight important for every ecosystem ?
Answer:

1. Sunlight is essential for the plants for photosynthesis.
2. It is the only source of energy for the entire ecosystem.
3. Without sunlight the food chains will not exist.
4. Survival of plants is therefore dependent on sunlight.
5. In case of animals diurnal and seasonal variations in light intensity and duration decide the feeding, foraging and reproductive activities.
6. Migrations shown by certain animals also depend on light.
7. Almost all animals have behaviour based on photoperiod. The proportion of sunlight on land also decides the ambient temperature. Thus, life is dependent on light.

Question 4.
What can be the causes of deviation from 1 : 1 sex ratio in natural habitat?
Answer:

1. In nature, when there is chromosomal sex determination, usually 1 : 1 sex ratio is obtained. But there are other causes for deviation from 1 : 1 sex ratio.
2. There are many other environmental sex determination methods where 1 : 1 sex
   ratio becomes skewed. This is due to mere chance of fertilization.
3. Also some lower animals show parthenogenesis e.g. as in honey bees. In such cases, offspring produced may not be in 1 : 1 proportion.

Question 5.
What are the special adaptations that endoparasites show?
Answer:

1. Endoparasites show loss of unnecessary sense organs as these are not needed for the parasite.
2. There are adhesive organs or suckers always present in the endoparasites which are needed to cling on to the host.
3. Endoparasites show loss of digestive system.
4. They have very high reproductive capacity.
5. The complex life cycles are seen in such parasites which involve intermediate hosts or vectors to facilitate transfer to the host.

Question 6.
What are the effects of parasite on the host?
Answer:

1. Most of the parasites cause harm to the host.
2. Host is affected by reducing its survival, growth and reproduction.
3. Some parasites can also be fatal to the host causing death of the host.
4. The population density of host species is reduced by parasites.
5. The host species become more vulnerable to predation by making it physically weak.

Question 7.
What is the role of predator in balancing the ecosystem?
Answer:

1. Predators keep prey population under control. If predators are lacking from the ecosystem, the prey population will rise without any control. Their high density may cause instability in ecosystem.
2. Predators also help in maintaining the species diversity in a community. This is done by reducing the intensity of competition among competing prey species.
3. Predators control the pest species and thus can be used for natural biological control measures in an ecosystem. E.g. frog controlling the locust population.
4. Predators also control the invading exotic species and stop their rapid spread of such species.

Question 8.
How does prey species defend themselves against predators?
Answer:
Prey species show following defence mechanisms:

1. Showing camouflage for concealment.
2. Moving at faster speed for escape.
3. Cryptic colouration to avoid the detection. This is seen in some insects. Also predators display such cryptic colouration to avoid detection. E.g. Colouration in frog.
4. Bad taste due to accumulated chemicals.
   E.g. The Monarch butterfly is highly distasteful to its predator bird as it stores a special chemical in the body during its caterpillar stage by feeding on a poisonous weed.

Question 9.
With suitable examples describe commensalism.
Answer:
I. Commensalism : Commensalism is the interaction between two species in which one species derives benefit and the other one is neither harmed nor benefited.
II. Examples of commensalism:

1. Orchid grows as epiphyte on other big trees. The tree do not get any benefit but is neither harmed. But orchid gets support.
2. Cattle egret is the insectivorous bird which forage close to cattle. When cattle move, the hidden insects in the grass are flushed out. These insects are then captured by egrets. Cattle do not get benefit but birds do.
3. Sea anemone has stinging cells on the tentacles which offer protection to clown fish. Clown fish gets the protection from other predators, whereas, sea anemone does not derive any benefit from this association.

Question 10.
What are different population attributes?
Answer:

1. Basic physical attributes of population are population size and population density.
2. Number of individuals in a population is its size whereas number of individuals present per unit space, in a given time is called its density.
3. The other attributes are natality or birth rate, mortality or death rate, immigration means coming into the population, emigration means leaving from the population, age pyramids, sex ratio and biotic potential etc.

Question 11.
What are the decisive factors for population density?
Answer:
The density of a population in a given habitat during a given period fluctuates due to changes in four basic processes, viz.

1. Natality i.e. birth rate (The number of births during a given period in the population that are added to the initial density).
2. Mortality i.e. death rate (The number of deaths in the population during a given period).
3. Immigration i.e. number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
4. Emigration i.e. the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.
5. Natality and immigration increase in population density whereas mortality and emigration decrease it.

Question 12.
Should an ideal parasite be able to thrive within the host without harming it?
Answer:
A parasite that resides inside the body of host, will certainly cause discomfort to the host. But if the host dies, the parasite also will diminish. Therefore, Ideal parasite will not extract more benefits from the host. But ideal parasite cannot stay alive without association with host. They need host for various purposes like nutrition, food. Some need the host for completing their life cycle. So parasite will always harm the host in order to thrive.

Question 13.
Why didn’t natural selection lead to the evolution of such totally harmless parasites?
Answer:
Being a parasite means causing trouble or harm to the host. When host-parasite relationship is concerned both of them evolve together which is also called co¬evolution. If totally harmless parasite has to be evolved through natural selection, then such parasite will not remain as a parasite but will become commensal or can show amensalism. Therefore, natural selection does not lead to evolution of totally harmless parasite.

Question 14.
What will happen when carrying capacity of any habitat is exceeded?
Answer:
When the carrying capacity of any habitat is exceeded there is severe resource crunch. This may be in the form of food, shelter or availability of mates. This results into struggle for survival. This will result into death of those who cannot cope up with the struggle. They may die due to starvation or due to interactions such as fierce competitions. This causes the population again to come back to equilibrium. Thus when carrying capacity is exceeded, the natural way is to reduce the population in the habitat.

Question 15.
What could be the reasons behind enormous increase in human population?
Answer:

1. Human population has increased enormously in last century due to increased natality and reduced mortality.
2. Due to advances in medical sciences, vaccinations and better life expectancy mortality rate has been considerably reduced. With the exception of Covid 19 Pandemic, all the mortalities due to epidemics were controlled due to modern medical facilities.
3. Due to better food production and advances in agriculture methods, people are not starving anymore. This has considerably reduced the deaths due to starvations.
4. However, natality rate has not been reduced especially in developing countries. Lack of family planning measures, ignorance, illiteracy and traditional thinking about birth of male child, all such factors have caused tremendous increase in the human population.

Question 16.
What can be reason behind the different reproductive strategies adopted by monocot plants like cereals / pulses and dicot plants like mango?
Answer:
The monocot plants such as cereals and pulses and also mango are commercially important cash crops. Therefore, in order to obtain profitable harvest, different reproductive strategies are adopted.

Chart based / Table based questions

Question 1.
Complete the following table by placing the categories and the given signs + : Benefited ; – : Inhibited ; 0 : Not affected

Interactions	Species
	A	B
1. Neutralism – …………………..
2. Negative interactions …………………………
3. Positive interactions …………………….
4. Both positive and negative interactions …………………………..

Answer:

Interactions	Species
	A	B
1. Neutralism – no significant effect	O	O
2. Negative interactions a. Competition – direct interference type b. Competition – resource – use type c. Amemsalism	– – –	– – O
3. Positive interactions a. Symbiosis (Mutualism) b. Commensalism c. Protocooperation	+ + +	+ O +
4. Both positive and negative interactions a. Parasitism b. Predation	+ +	– –

Diagram based questions

Question 1.
Sketch and label the graph showing logistic growth curve of the population.
Answer:

Question 2.
Sketch and label the graph showing exponential growth curve of the population.
Answer:

Question 3.
Sketch the different types of Age Pyramids.
Answer:

Question 4.
From the age pyramids given in figure what will be your forecast for 15 years from now for the populations of 1. Kenya, 2. Australia, 3. Italy and 4. Hungary?

Answer:
(1) Forecast for Kenya : Kenya is showing rapid growth as its pre-reproductive group is large. Post-reproductive group is much reduced. Therefore, 15 years from now, the population will be large.

(2) Forecast for Australia : Australian age pyramid is showing larger post-reproductive group. In the years to come, they will be removed from the population. Also the pre-reproductive group is not much large. This will not add to the Australian population in the future. Thus, this country will show slow growth.

(3) Forecast for Italy : In Italy the reproductive and post-reproductive groups are almost similar. The pre-reproductive group is also limited. Hence, in future, the population will not expand. At the same time the post-reproductive group here is large which means that the population growth will be reduced in the next 15 years.

(4) Forecast for Hungary : Hungary is showing : typical age pyramid where the pre-reproductive and reproductive groups are small. On the contrary, the post-reproductive group is more, which shows that in the next 15 years, the old people will leave the population, causing negative growth.

Long answer questions

Question 1.
What are the characteristics of ecological niche ?
Answer:

1. A niche describes how that organism is linked with its physical and biological environment.
2. Niche is described as a position of a species in the environment. It gives the idea about how the organisms are surviving and fulfilling their needs of shelter and food.
3. By studying niche one can get idea of the flow of energy from one organism to another. This helps to understand the feeding habits and interactions involving food chain and food web.
4. If any niche is left vacant, other organisms\fill that position.
5. The niche is specific to each species. Two species can never share the same niche. By having specific niche, every organism tries to reduce competition for resources.
6. E.g. In birds, each one is specific in their eating habits, some are insectivorous, while some are frugivorous. Some are omnivorous, in this way birds living in the same habitat differ in their niches because of different eating habits.

Question 2.
What are the different ways in which organisms adapt to the changes in the environment?
Answer:
To survive and propagate further in any environment, organisms show one of the four possible ways, viz. regulate, conform, migrate and suspend.
(1) Regulate : In this method, organisms maintain homeostasis by physiological and behavioural changes. Due to homeostatic regulation, they can perform thermoregulation or osmoregulation. E.g. All birds and mammals show constant body temperature and osmotic concentration irrespective of external temperature.

(2) Conform : Most of the animals and plants are unable to maintain a constant internal environment. Their body parameters change according to outside environment. E.g. Poikilothermic animals cannot maintain body temperature but they are simple conformers. In few aquatic animals, the osmotic concentration of the body fluids changes according to surrounding osmotic concentration. Few conformers can regulate the parameters in limited range.

(3) Migrate : When organism is unable to cope up with surrounding temperatures, they migrate temporarily from such stressful habitat to a more favourable habitat. After the stressful period is over, they return back. Birds show long-distance migrations during severe winter.

(4) Suspend : Suspending the life activities for particular period is one of the methods to cope up with stressful conditions. Seeds of plants remain dormant over unfavourable period and once favourable conditions are resumed they start growing. This state is called dormancy during which metabolic activities are suspended. Hibernation and aestivation seen in some animals is also for escaping severe winter or summer respectively. E.g. Polar bear shows hibernation while snails and fish show aestivation. These are also suspension measures.

Question 3.
What are the adaptations in animals living under crushing pressure at great depths of ocean?
Answer:

1. Environment of depths of ocean are characterized by high pressure, low temperature, absence of light, calmness of water, absence of phytoplankton and other producers, scarcity of food and thus animals staying here show many adaptations.
2. Due to extreme pressure the bodies of deep- sea fish and other animals are very much compressed.
3. The bony skeletons are much reduced except for jaws. They have watery muscles. Some deep-sea fishes exhibit greatly enlarged eyes which act like telescope.
4. They are highly effective as in depths there is less light. Their retina is composed of a number of tiers of rods, presumably arranged to absorb all the limited light that enters the eye. However, the eye-size is small.
5. Some benthic fishes have eyes located on only one side of the body. E.g. Sole fish.
6. Many deep-sea animals are bioluminescent, i.e. they produce their own light by means of luminous organs.
7. In anglerfish, the light is used as a bait to attract prey and also for species and sex recognition.
8. The mouth of deep-sea fish is the enormous, which enables them to gulp large sized prey.
9. Many of the deep-sea animals have long appendages, abundant spines, stalks or other

Question 4.
With the help of suitable diagram describe the Exponential population growth curve.
Answer:

1. When the resources are abundant, organisms show continuous growth of a population without any hindrance. With unlimited resources, each species has the ability to realise fully its innate potential to grow in number.
2. Such growth of a population is called an exponential or geometric growth.
3. Any species growing exponentially under unlimited resource conditions can reach enormous population densities in a short time. E.g. Human population shows such exponential growth.
4. Exponential growth shows J-shaped curve.

Question 5.
Explain the ‘Competitive Exclusion Principle’ given by Gause.
Answer:
Gause’s ‘Competitive Exclusion Principle’:
(1) This principle states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually.

(2) The Gause’s principle may be true if resources are limiting, but not otherwise. More recent studies do not support such gross generalisations about competition. The species during competition also show resource partitioning.

(3) In resource partitioning, the species facing competition might evolve mechanisms that promote co-existence rather than exclusion. If two species compete for the same resource, they could avoid competition by choosing, for instance, different times for feeding or different foraging patterns. E.g. Five closely related species of warblers living on the same tree were able to avoid competition and co-exist due to behavioural differences in their foraging activities. If there are two competing species and one is comparatively superior than the other, then the inferior one remains restricted to smaller geographical area. If this superior species is removed then only the inferior species expands its range.

Question 6.
What are the key differences that make such a great variation in the physical and chemical conditions of different habitats?
Answer:
Different abiotic factors make great variations in the physical and chemical conditions of different habitats.
The most important abiotic factors are temperature, water, light and soil. These abiotic factors act with the resident biotic factors in that habitat. Biotic components such as pathogens, parasites, predators and competitors keep on interacting continuously. All such interactions cause variations in different habitats.

Question 7.
Give names of eurythermal and stenothermal animals and plants?
Answer:

1. Eurythermal animals : Goat, man, cow, crab, bivalves, etc.
2. Stenothermal animals : Insects, reptiles, snakes, fishes, etc.
3. Eurythermal plants : Roses, daisies, oak trees, some fruits and vegetables.
4. Stenothermal plants : Croton, Bougainvillea, Frangipani, vines and orchids, some other fruits and vegetables

Question 8.
What will be the effect of increasing global temperatures on the different habitats and the organisms found in those habitats?
Answer:

1. Global temperature rise or global warming is having great impact on natural habitats.
2. Aquatic habitats like oceans are worst affected as 90% of extra heat enters marine waters. These elevated temperatures cause adverse effects on marine habitat.
3. Coral reefs are also affected due to increased temperature. It causes bleaching of coral reefs. Fish populations are adversely affected.
4. Due to rising temperature, the tundra regions and polar regions are showing melting of snow. These habitats are disappearing and the animals from these areas such as polar bears are on the verge of extinction. Increased temperatures also cause desertification of the once green habitats.
5. Nearly 50% of the species are under threat of extinction due to climate change. Global warming and climate change thus reduces biodiversity.
6. Every plant and animal species is adapted to a specific temperature conditions. But due to global warming and associated climate change, these species are affected. Some species show migrations to cooler places.
7. Temperatures also alter the life cycles of plants and animals. When temperatures rise, many plants grow rapidly and bloom earlier in the spring and survive longer into the fall. Some animals leave hibernation sooner.

Question 9.
Give examples of an animal and plant that can survive in fresh water as well as sea water.
Answer:
Few fish species such as salmons, crustacean prawns and crabs are able to survive in fresh water as well as sea water. Among plants, few phytoplankton, some algal species and mangorves show similar phenomena.

Question 10.
What is the source of energy for the life in deep ocean trenches where sunlight does not reach?
Answer:
In deep ocean trenches, the sunlight does not penetrate. Hence photosynthesis is not possible here. Most of the organisms that live here depend upon subsistence on falling organic matter. This organic matter falls off down from the upper photic zones, where phytoplankton can perform photosynthesis. This organic matter acts as source of energy for the life in deep oceanic trenches.

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Question 1.
Define chemistry.
Answer:
Chemistry is the study of matter, its physical and chemical properties and the physical and chemical changes it undergoes under different conditions.

Question 2.
Why is chemistry called a central science?
Answer:

1. Knowledge of chemistry is required in the studies of physics, biological sciences, applied sciences, and earth and space sciences.
2. Chemistry is involved in every aspect of day-to-day life, i.e. the air we breathe, the food we eat, the fluids we drink, our clothing, transportation and fuel supplies, etc.

Hence, chemistry is called a central science.

Question 3.
Give reason: Although chemistry has ancient roots, it has developed as a modern science.
Answer:
Technological development in sophisticated instruments have expanded knowledge of chemistry which, now, has been used in applied sciences such as medicine, dentistry, engineering, agriculture and in daily home use products. Hence, due to development and advancement in science and technology, chemistry has developed as modem science.

Question 4.
How is chemistry traditionally classified?
Answer:
Chemistry is traditionally classified into five branches:

• Organic chemistry
• Inorganic chemistry
• Physical chemistry
• Biochemistry
• Analytical chemistry

Question 5.
Explain the following terms:
i. Organic chemistry
ii. Inorganic chemistry
iii. Physical chemistry
Answer:
i. Organic chemistry: It deals with properties and reactions of compounds of carbon.
ii. Inorganic chemistry: It deals with the study of all the compounds which are not organic.
iii. Physical chemistry: It deals with the study of properties of matter, the energy changes and the theories, laws and principles that explain the transformation of matter from one form to another. It also provides basic framework for all the other branches of chemistry.

Question 6.
Distinguish between
i. Mixtures and pure substances
ii. Mixtures and compounds
Answer:
i.

	Mixtures	Pure substances
a.	Mixtures have no definite chemical composition.	Pure substances have a definite chemical composition.
b.	Mixtures have no definite properties.	Pure substances always have the same properties regardless of their origin.
e.g.	Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.	Pure metal, distilled water, etc.

ii.

	Mixtures	Compounds
a.	Mixtures have no definite chemical composition.	Compounds are made up of two or more elements in fixed proportion.
b.	The constituents of a mixture can be easily separated by physical method.	The constituents of a compound cannot be easily separated by physical method.
e.g.	Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.	Water, table salt, sugar, etc.

Question 7.
What is the difference between element and compound?
Answer:
Elements cannot be broken down into simpler substances while compounds can be broken down into simpler substances by chemical changes.

Question 8.
Explain: States of matter
Answer:
There are three different states of matter as follows:

1. Solid: Particles are held tightly in perfect order. They have definite shape and volume.
2. Liquid: Particles are close to each other but can move around within the liquid.
3. Gas: Particles are far apart as compared to that of solid and liquid.

These three states of matter can be interconverted by changing the conditions of temperature and pressure.

Question 9.
Explain: Physical and chemical properties
Answer:
i. Physical properties: These are properties which can be measured or observed without changing the identity or the composition of the substance. e.g. Colour, odour, melting point, boiling point, density, etc.

ii. Chemical properties: These are properties in which substances undergo change in chemical composition. e.g. Coal bums in air to produce carbon dioxide, magnesium wire bums in air in the presence of oxygen to form magnesium oxide, etc.

Question 10.
How are properties of matter measured?
Answer:

• Measurement involves comparing a property of matter with some fixed standard which is reproducible and unchanging.
• Properties such as mass, length, area, volume, time, etc. are quantitative in nature and can be measured.
• A quantitative measurement is represented by a number followed by units in which it is measured.
• These units are arbitrarily chosen on the basis of universally accepted standards. e.g. Length of class room can be expressed as 10 m. Here, 10 is the number and ‘m’ is the unit ‘metre’ in which the length is measured.

Question 11.
Define: Units
Answer:
The arbitrarily decided and universally accepted standards are called units.
e.g. Metre (m), kilogram (kg).

Question 12.
What are the various systems in which units are expressed?
Answer:
Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.

Question 13.
What are SI units? Name the fundamental SI units.
Answer:
SI Units: In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI system (abbreviated from its French name).
The seven fundamental SI units are as given below:

No.	Base physical quantity	SI unit	Symbol
i.	Length	Metre	k
ii.	Mass	Kilogram	kg
iii.	Time	Second	s
iv.	Temperature	Kelvin	K
v.	Amount of substance	Mole	mol
vi.	Electric current	Ampere	A
vii.	Luminous intensity	Candela	cd

[Note: Units for other quantities such as speed, volume, density, etc. can be derived from fundamental SI units.]

Question 14.
What is the basic unit of mass in the SI system?
Answer:
The basic unit of mass in the SI system is kilogram (kg).

Question 15.
Name the following:
i. Full form of CGS unit system
ii. Full form of FPS unit system
iii. The SI unit of length
iv. Symbol used for Candela unit
v. SI unit of electric current
vi. SI unit of electric current
Answer:
i. Centimetre Gram Second
ii. Foot Pound Second
iii. Metre (m)
iv. Cd
v. Kelvin (K)
vi. Ampere (A)

Question 16.
Give reason: The mass of a body is more fundamental property than its weight.
Answer:

• Mass is an inherent property of matter and is the measure of the quantity of matter of a body.
• The mass of a body does not vary with respect to its position.
• On the other hand, the weight of a body is a result of the mass and gravitational attraction
• Weight varies because the gravitational attraction of the earth for a body varies with the distance from the centre of the earth.

Hence, the mass of a body is more fundamental property than its weight.

Question 17.
How is gram related to the SI unit kilogram?
Answer:
The SI unit kilogram (kg) is related to gram (g) as 1 kg = 1000 g= 10³ g.
[Note: ‘Gram’ is used for weighing small quantities of chemicals in the laboratories.
Other commonly used quantity is ‘milligram’. 1 kg = 1000 g = 10⁶ mg]

Question 18.
Why are fractional units of the SI units of length often used? Give two examples of the fractional units of length. How are they related to the SI unit of length?
Answer:
i. Some properties such as the atomic radius, bond length, wavelength of electromagnetic radiation, etc. are very small and therefore, fractional units of the SI unit of length are often used to express these properties.
ii. Fractional units of length: Nanometre (nm), picometre (pm), etc.
iii. Nanometre (nm) and picometre (pm) are related to the SI unit of length (m) as follows:
1 nm = 10^-9 m, 1 pm = 10^-12 m

Question 19.
Define: Volume
Answer:
Volume is the amount of space occupied by a three-dimensional object. It does not depend on shape.

Question 20.
State the common unit used for the measurement of volume of liquids and gases.
Answer:
The common unit used for the measurement of volume of liquids and gases is litre (L).

Question 21.
How is the SI unit of volume expressed?
Answer:
The SI unit of volume is expressed as (metre)³ or m³.

Question 22.
Name some glassware that are used to measure the volume of liquids and solutions.
Answer:

• Graduated cylinder
• Burette
• Pipette

Question 23.
What is a volumetric flask used for in laboratory?
Answer:
A volumetric flask is used to prepare a known volume of a solution in laboratory.

Question 24.
What is density of a substance? How is it measured?
Answer:
Density:

• Density of a substance is its mass per unit volume. It is the characteristic property of any substance.
• It is determined in the laboratory by measuring both the mass and the volume of a sample.
• The density is calculated by dividing mass by volume.

Question 25.
How is the SI unit of density derived? State CGS unit of density.
Answer:
i. The SI unit of density is derived as follows:
Density = \(\frac{\text { SI unit mass }}{\text { SI unit volume }}\)
= \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
= kg m^-3

ii. CGS unit of density: g cm^-3
[Note: The CGS unit, g cm^-3 is equivalent to \(\frac{\mathrm{g}}{\mathrm{mL}}\) or g mL^-1.]

Question 26.
State three common scales of temperature measurement.
Answer:

1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)

Question 27.
State the temperatures in Fahrenheit scale that corresponds to 0 °C and 100 °C.
Answer:
The temperature that corresponds to 0 °C is 32 °F and the temperature that corresponds to 100 °C is 212 °F.

Question 28.
Write the expression showing the relationship between:
i. Degree Fahrenheit and Degree Celsius
ii. Kelvin and Degree Celsius
Answer:
i. The relationship between degree Fahrenheit and degree Celsius is expressed as,
°F = \(\frac {9}{5}\) (°C) + 32
ii. The relationship between Kelvin and degree Celsius is expressed as,
K = °C + 273.15

Question 29.
Convert the following degree Fahrenheit temperature to degree Celsius.
i. 50 °F ii. 10 °F
Answer:
Given: Temperature in degree Fahrenheit = 50 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 50 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
50 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(50-32) \times 5}{9}\)
= 10 °C

ii. Given: Temperature in degree Fahrenheit = 10 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 10 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
10 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(10-32) \times 5}{9}\)
= -12.2 °C
Ans: i. The temperature 50 °F corresponds to 10 °C.
ii. The temperature 10 °F corresponds to -12.2 °C.

Question 30.
What is a chemical combination?
Answer:

• The process in which the elements combine with each other to form compounds is called chemical combination.
• The process of chemical combination is governed by five basic laws which were discovered before the knowledge of molecular formulae.

Question 31.
State and explain the law of definite proportions.
Answer:
Law of definite proportions:
i. The law states that “A given compound always contains exactly the same proportion of elements by weight”.
ii. French chemist, Joseph Proust worked with two samples of cupric carbonate; one of which was naturally occurring cupric carbonate and other was synthetic sample. He found the composition of elements present in both the samples was same as shown below:

Cupric carbonate	% of copper	% of carbon	% of oxygen
Natural sample	51.35	9.74	38.91
Synthetic sample	51.35	9.74	38.91

iii. Thus, irrespective of the source, a given compound always contains same elements in the same proportion.

Question 32.
State and explain the law of multiple proportions.
Answer:
Law of multiple proportions:
i. John Dalton (British scientist) proposed the law of multiple proportions in 1803.
ii. It has been observed that two or more elements may combine to form more than one compound.
iii. The law states that, “ When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.
e.g. Hydrogen and oxygen combine to form two compounds, water and hydrogen peroxide.

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.

Question 33.
Show that NO and NO₂ satisfy the law of multiple proportions.
Answer:
Nitrogen and oxygen combine to form two compounds, nitric oxide (NO) and nitrogen dioxide (NO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of nitrogen (14 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 34.
Show that carbon monoxide and carbon dioxide satisfy the law of multiple proportions.
Answer:
Chemical reaction of carbon with oxygen gives two compounds, carbon monoxide (CO) and carbon dioxide (CO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of carbon (12 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 35.
Show that SO₂ and SO₃ satisfy the law of multiple proportions.
Answer:
Chemical reaction of sulphur with oxygen gives two compounds, sulphur dioxide (SO₂) and sulphur trioxide (SO₃).

Here, the two masses of oxygen (32 g and 48 g) which combine with the fixed mass of sulphur (32 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 32 : 48 or 2 : 3.
This is in accordance with the law of multiple proportions.

Question 36.
State and explain Gay Lussac’s law of gaseous volume with two examples.
Answer:
Gay Lussac’s law:
i. Gay Lussac proposed the law of gaseous volume in 1808.
ii. Gay Lussac’s law states that, “ When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
e.g. a. Under identical conditions of temperature and pressure, 100 mL of hydrogen gas combine with 50 mL of oxygen gas to produce 100 mL of water vapour.

Thus, the simple ratio of volumes is 2 : 1 : 2.

b. Under identical conditions of temperature and pressure, 1 L of nitrogen gas combine with 3 L of hydrogen gas to produce 2 L of ammonia gas.

Thus, the simple ratio of volumes is 1 : 3 : 2.

Question 37.
Give two examples which support the Gay Lussac’s law of gaseous volume.
Answer:
i. Under identical conditions of temperature and pressure, 1 L of hydrogen gas reacts with 1 L of chlorine gas to produce 2 L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2
This is in accordance with Gay Lussac’s law.

ii. Under identical conditions of temperature and pressure, 200 mL sulphur dioxide combine with 100 mL oxygen to form 200 mL sulphur trioxide.

Thus, the ratio of volumes is 2 : 1 : 2.
This is in accordance with Gay Lussac’s law.

Question 38.
Match the following:

	Law		Statement
i.	Law of definite proportions	a.	When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
ii.	Gay Lussac’s law	b.	Equal volumes of all gases at the same temperature and pressure contain equal number of molecules
iii.	Law of multiple proportions	c.	When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
iv.	Avogadro’s law	d.	A given compound always contains exactly the same proportion of elements by weight

Answer:
i – d,
ii – c,
iii – a,
iv – b

Question 39.
32 g of oxygen reacts with some carbon to make 56 grams of carbon monoxide. Find out how much mass must have been used.
Answer:
Given: Mass of oxygen (reactant) = 32 g, mass of carbon monoxide (product) = 56 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 16 g oxygen to form 28 g of carbon monoxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 16 = 32 g) of oxygen to give (2 × 28 = 56 g) carbon monoxide.
Ans: Mass of carbon used = 24 g

Question 40.
Calculate the mass of sulphur trioxide produced by burning 64 g of sulphur in excess of oxygen. (Average atomic mass: S = 32 u, O = 16 u).
Solution:
Given: Mass of sulphur (reactant) = 64 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 48 g oxygen to form 80 g of sulphur trioxide as follows:

Hence, (2 × 32 = 64 g) of sulphur will combine with (2 × 48 = 96 g) of oxygen to give (2 × 80 = 160 g) sulphur trioxide.
Ans: Mass of sulphur trioxide produced = 160 g

Question 41.
Explain Dalton’s atomic theory.
Answer:
John Dalton published “A New System of chemical philosophy” in the year of 1808. He proposed the following features, which later became famous as Dalton’s atomic theory.

• Matter consists of tiny, indivisible particles called atoms.
• All the atoms of a given elements have identical properties including mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.

Dalton’s atomic theory could explain all the laws of chemical combination.

Question 42.
Give reason: Dalton’s atomic theory explains the law of conservation of mass.
Answer:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• According to Dalton’s atomic theory, chemical reactions involve only the reorganization of atoms. Therefore, the total number of atoms in the reactants and products should be same and mass is conserved during a reaction.

Hence, Dalton’s atomic theory explains the law of conservation of mass.

Question 43.
Give reason: Dalton’s atomic theory explains the law of multiple proportion.
Answer:

• The law of multiple proportion states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
• According to Dalton’s atomic theory, compounds are formed when atoms of different elements combine in fixed ratio.

Hence, Dalton’s atomic theory explains the law of multiple proportion.

Question 44.
Define: Atomic mass unit (amu).
Answer:
Atomic mass unit or amu is defined as a mass exactly equal to one twelth of the mass of one carbon-12 atom.

Question 45.
How is relative atomic mass of an atom measured?
Answer:

• The mass of a single atom is extremely small, i.e. the mass of a hydrogen atom is 1.6736 × 10^-24 g. Hence, it is not possible to weigh a single atom.
• In the present system, mass of an atom is determined relative to the mass of an atom of carbon-12 as the standard. This was decided in 1961 by international agreement.
• The atomic mass of carbon-12 is assigned as 12.00000 atomic mass unit (amu).
• The masses of all other elements are determined relative to the mass of an atom of carbon-12 (C-12).
• The atomic masses are expressed in amu which is exactly equal to one twelth of the mass of one carbon-12 atom.
• The value of 1 amu is equal to 1.6605 × 10^-24 g.

Question 46.
What is meant by Unified Mass unit?
Answer:

• Presently, instead of amu, Unified Mass has now been accepted as the unit of atomic mass.
• It is called Dalton and its symbol is ‘u’ or ‘Da’.

Question 47.
What is average atomic mass?
Answer:
The atomic mass of an element which exists as mixture of two or more isotopes is the average of atomic masses of its isotopes. This is called average atomic mass.

Question 48.
Define: Molecular mass
Answer:
Molecular mass of a substance is the sum of average atomic masses of the atoms of the elements which constitute the molecule.
OR
Molecular mass of a substance is the mass of one molecule of that substance relative to the mass of one carbon-12 atom.

Question 49.
How is molecular mass of a substance calculated? Give example.
Answer:
Molecular mass is calculated by multiplying average atomic mass of each element by the number of its atoms and adding them together.
e.g. Molecular mass of carbon dioxide (CO₂) is calculated as follows:
Molecular mass of CO₂ = (1 × average atomic mass of C) + (2 × average atomic mass of O)
= (1 × 12.0 u) + (2 × 16.0 u)
= 44.0 u

Question 50.
Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g. What is the atomic mass of hydrogen in u?
Solution:
Given: Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g.
To find: Atomic mass of hydrogen in u
Calculation: 1.66056 × 10^-24 g = 1 u
∴ 1.6736 × 10^-24 g = x
x = \(\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g} / \mathrm{u}}\) = 1.008u
Ans: The atomic mass of hydrogen in u = 1.008 u

Question 51.
The mass of an atom of one carbon atom is 12.011 u. What is the mass of 20 atoms of the same isotope?
Solution:
Mass of l atom of carbon = 12.011 u
∴ Mass of 20 atoms of same carbon isotope = 20 × 12.011 u = 240.220 u
Ans: The mass of 20 atoms of same carbon isotope = 240.220 u

Question 52.
Calculate the average atomic mass of neon using the following data:

Isotope	Atomic mass	Natural Abundance
20Ne	19.9924 u	90.92%
21Ne	20.9940 u	0.26 %
22Ne	21.9914 u	8.82 %

Solution:
Average atomic mass of Neon (Ne)

Ans: Average atomic mass of neon = 20.1707 u

Question 53.
Calculate the average atomic mass of argon from the following data:

Isotope	Isotopic mass (g mol-1)	Abundance
36Ar	35.96755	0.337%
38Ar	37.96272	0.063%
40Ar	39.9624	99.600%

Solution:
Average atomic mass of argon (Ar)

Ans: Average atomic mass of argon = 39.974 g mol^-1

Question 54.
Calculate the molecular mass of the following in u:
i. H₂O ii. C₆H₅Cl iii. H₂SO₄
Solution:
i. Molecular mass of H₂O = (2 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 1.0u) + (1 × 16.0 u)
= 18 u

ii. Molecular mass of C₆H₅Cl = (6 × Average atomic mass of C) + (5 × Average atomic mass of H) + (1 × Average atomic mass of Cl)
= (6 × 12.0 u) + (5 × 1.0 u) + (1 × 35.5 u)
= 112.5 u

iii. Molecular mass of H₂SO₄ = (2 × Average atomic mass of H) + (1 × Average atomic mass of S) + (4 × Average atomic mass of O)
= (2 × 1.0 u) + (1 × 32.0 u) + (1 × 16.0 u)
= 98 u
Ans: i. The molecular mass of H₂O = 18 u
ii. The molecular mass of C₆H₅Cl = 112.5 u
iii. The molecular mass of H₂SO₄ = 98 u

Question 55.
Find the mass of 1 molecule of oxygen (O₂) in amu (u) and in grams.
Solution:
Molecular mass of O₂ = 2 × 16 u
∴ Mass of 1 molecule = 32 u
∴ Mass of 1 molecule of O₂= 32 × 1.66056 × 10^-24 g = 53.1379 × 10^-24 g
Ans: Mass of 1 molecule in amu = 32 u
Mass of 1 molecule in grams = 53.1379 × 10^-24 g

Question 56.
Find the formula mass of
i. NaCl ii. Cu(NO₃)₂
Solution:
i. Formula mass of NaCl
= Average atomic mass of Na + Average atomic mass of Cl
= 23.0 u + 35.5 u = 58.5 u

ii. Formula mass of Cu(NO₃)₂
= Average atomic mass of Cu + 2 × (Average atomic mass of N + Average atomic mass of three O)
= 63.5 + 2 × [14 + (3 × 16)] = 187.5 u
Ans: i. Formula mass of NaCl = 58.5 u
ii. Formula mass of Cu(NO₃)₂ = 187.5 u

Question 57.
Find the formula mass of
i. KCl
ii. AgCl
Atomic mass of K = 39 u, Ag =108 u and Cl = 35.5 u.
Solution:
i. Formula mass of KCl
= Average atomic mass of K + Average atomic mass of Cl
= 39 u + 35.5 u = 74.5 u

ii. Formula mass of AgCl
= Average atomic mass of Ag + Average atomic mass of Cl
= 108 + 35.5 = 143.5 u
Ans: i. Formula mass of KCl = 74.5 u
ii. Formula mass of AgCl = 143.5 u

Question 58.
Calculate the number of moles and molecules of urea present in 5.6 g of urea.
Solution:
Given: Mass of urea = 5.6 g
To find: The number of moles and molecules of urea
Formulae: i. Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Mass of urea = 5.6 g
Molecular mass of urea, NH₂CONH₂
= (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O)
= (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) = 60 u
∴ Molar mass of urea = 60 g mol^-1
∴ Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{5.6 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.09333 mol

[Calculation using log table:
\(\frac{5.6}{60}\)
= Antilog₁₀ [log₁₀ (5.6) – log₁₀ (60)]
= Antilog₁₀ [0.7482 – 1.7782]
= Antilog₁₀ [latex]\overline{2} .9700[/latex]
= 0.09333]

Now,
Number of molecules of urea
= Number of moles × Avogadro’s constant
= 0.09333 mol × 6.022 × 10²³ molecules/mol
= 0.5616 × 10²³ molecules (by using log table)
= 5.616 × 10²² molecules
Ans: Number of moles of urea = 0.0933 mol
Number of molecules of urea = 5.616 × 10²² molecules

[Calculation using log table:
0.09333 × 6.022
= Antilog₁₀ [log₁₀ (0.09333) + log₁₀ (6.022)]
= Antilog₁₀ [\(\overline{2} .9698\) + 0.7797]
= Antilog₁₀ [latex]\overline{1} .7495[/latex]
= 0.5616]

Question 59.
Calculate the number of atoms in each of the following:
i. 64 u of oxygen (O)
ii. 42 g of nitrogen (N)
Solution:
i. 64 u of oxygen (O) = x atoms
Atomic mass of oxygen (O) = 16 u
∴ Mass of one oxygen atom = 16 u
∴ x = \(\frac{64 \mathrm{u}}{16 \mathrm{u}}\) = 4 atoms

ii. 42 g of nitrogen (N)
Atomic mass of nitrogen = 14 u
∴ Molar mass of nitrogen = 14 g mol^-1
Now,
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{42 \mathrm{~g}}{14 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3 mol
Now,
Number of atoms = Number of moles × Avogadro’s constant
= 3 mol × 6.022 × 10²³ atoms/mol
= 18.07 × 10²³ atoms
= 1.807 × 10²⁴ atoms
Ans: i. Number of oxygen atoms in 64 u = 4 atoms
ii. Number of nitrogen atoms in 42 g = 1.807 × 10²⁴ atoms

Question 60.
Calculate the number of atoms in each of the following.
i. 52 moles of Argon (Ar)
ii. 52 u of Helium (He)
iii. 52 g of Helium (He)
Solution:
i. 52 moles of Argon
1 mole Argon atoms = 6.022 × 10²³ atoms of Ar
∴ Number of atoms = 52 mol × 6.022 × 10²³ atoms/mol
= 313.144 × 10²³ atoms of Argon

ii. 52 g of He
Molar mass of He = mass of 1 atom of He = 4.0 u
4.0 u = 1 He
∴ 52 u = x
∴ x = 52 u × \(\frac{1 \text { atom of He }}{4.0 \mathrm{u}}\) = 13 atoms of He

iii. 52 g of He
Molar mass of He = 4.0 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{52 \mathrm{~g}}{4.0 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 13 mol
Number of atoms of He = Number of moles × Avogadro’s constant
= 13 mol × 6.022 × 10²³ atoms/mol
= 78.286 × 10²³ atoms of He
Ans. i. Number of argon atoms in 52 moles = 313.144 × 10²³ atoms of Argon
ii. Number of helium atoms in 52 u = 13 atoms of He
iii. Number of helium atoms in 52 g = 78.286 × 10²³ atoms of He

Question 61.
Calculate the number of atoms of ‘C’, ‘H’ and ‘O’ in 72.5 g of isopropanol, C₃H₇OH (molar mass 60 g mol^-1).
Solution:
Mass of isopropanol(C₃H₇OH) = 72.5 g
The number of atoms of C, H, O
Calculation: Molecular formula of isopropanol, is C₃H₇OH.
Molar mass of C₃H₇OH = 60 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{72.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 1.208 mol
∴ Moles of isopropanol = 1.21 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of isopropanol has total 12 atoms, out of which 8 atoms are of H, 3 of C and 1 of O.
∴ Number of C atoms in 72.5 g isopropanol = (3 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 2.182 × 10²⁴ atoms of carbon
∴ Number of ‘H’ atoms in 72.5 g isopropanol = (8 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 5.819 × 10²⁴ atoms of hydrogen
∴ Number of ‘O’ atoms in 72.5 g isopropanol = (1 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 7.274 × 10²³ atoms of oxygen
Ans. 72.5g of isopropanol contain 2.182 ×10²⁴ atoms of H and 7.274 × 10²³ atoms of O.

Question 62.
Calculate the number of moles and molecules of ammonia (NH₃) gas in a volume 67.2 dm³ of it measured at STP.
Solution:
Given: Volume of ammonia at STP = 67.2 dm³
To find: Number of moles and molecules of ammonia
Formulae: i. Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
ii. Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Number of moles (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Number of moles of NH₃ = \(\frac{67.2 \mathrm{dm}^{3}}{22.4 \mathrm{dm}^{3} \mathrm{~mol}^{-1}}\)
Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
3.0 mol × 6022 × 10²³ molecules mol^-1
= 18.066 × 10²³ molecules
Ans: Number of moles of ammonia = 3.0 mol
Number of molecules of ammonia = 18.066 × 10²³ molecules

Question 63.
3.40 g of ammonia at STP occupies volume of 4.48 dm³. Calculate molar mass of ammonia.
Solution:
Given: Mass of ammonia = 3.40 g
Volume at STP = 4.48 dm³
To Find: Molar mass of ammonia
Calculation: Let ‘x’ grams be the molar mass of NH₃.
Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Volume occupied by 3.40 g of NH₃ at S.T.P = 4.48 dm³
Volume occupied by ‘x’ g of NH3 at S.T.P = 22.4 dm³
∴ x = \(\frac{22.4 \times 3.40}{4.48}\) = 17.0 g mol^-1
Ans: Molar mass of ammonia is 17.0 g mol^-1.

Question 64.
Veg puffs from a particular bakery have an average mass of 27.0 g, whereas egg puffs from the same bakery have an average mass of 40 g.
i. Suppose a person buys 1 kg of veg puff from the bakery. Calculate the number of veg puffs he receives.
ii. Determine the mass of egg puffs (in kg) that will contain the same number of eggs puffs as in one kilogram of veg puffs.
Solution:
i. Mass of a veg puff = 27.0 g = 0.027 kg
∴ Number of veg puffs in 1 kg = 1 / 0.027 = 37
ii. One kilogram of veg puffs contains 37 veg puffs.
Mass of 37 egg puffs = 37 × 0.040 = 1.48 kg
Ans: i. 37 veg puffs in 1 kg of puff.
ii. Mass of 37 egg puffs is 1.48 kg

Multiple Choice Questions:

1. The branch of chemistry which deals with carbon compounds is called ……………. chemistry.
(A) organic
(B) inorganic
(C) carbon
(D) bio
Answer:
(A) organic

2. A/an is a simple combination of two or more substances in which the constituent substances retain their separate identities.
(A) compound
(B) mixture
(C) element
(D) All of these
Answer:
(B) mixture

3. Which one of the following is NOT a mixture?
(A) Paint
(B) Gasoline
(C) Liquefied Petroleum Gas (LPG)
(D) Distilled water
Answer:
(D) Distilled water

4. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with ………………
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of reciprocal proportion
Answer:
(C) law of conservation of mass

5. A sample of calcium carbonate (CaCO₃) has the following percentage composition: Ca = 40 %; C = 12 %; O = 48 %. If the law of definite proportions is true, then the weight of calcium in 4 g of a sample of calcium
carbonate from another source will be ……………..
(A) 0.016 g
(B) 0.16 g
(C) 1.6 g
(D) 16 g
Answer:
(C) 1.6 g

6. Two elements, A and B, combine to form two compounds in which ‘a’ g of A combines with ‘b₁’ and ‘b₂’g of B respectively. According to law of multiple proportion ………………
(A) b₁ = b₂
(B) b₁ and b₂ bear a simple whole number ratio
(C) a and b₁ bear a whole number ratio
(D) no relation exists between b₁ and b₂
Answer:
(B) b₁ and b₂ bear a simple whole number ratio

7. At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour. This is in accordance with ……………….
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of gaseous volumes
Answer:
(D) law of gaseous volumes

8. One mole of oxygen molecule weighs …………….
(A) 8 g
(B) 32 g
(C) 16 g
(D) 6.022 × 10²³ g
Answer:
(B) 32 g

9. The mass of 0.002 mol of glucose (C₆H₁₂O₆) is ………………
(A) 0.20 g
(B) 0.36 g
(C) 0.50 g
(D) 1.80 g
Answer:
(B) 0.36 g

10. Which of the following is CORRECT?
(A) 1 mole of oxygen atoms contains 6.0221367 × 10²³ atoms of oxygen.
(B) 1 mole of water molecules contains 6.0221367 × 10²³ molecules of water.
(C) 1 mole of sodium chloride contains 6.0221367 × 10²³ formula units of NaCl.
(D) All of these
Answer:
(D) All of these

11. 180 g of glucose (C₆H₁₂O₆) contains ……………. carbon atoms.
(A) 1.8 × 10²³
(B) 1.8 × 10²⁴
(C) 3.6 × 10²³
(D) 3.6 × 10²⁴
Answer:
(C) 3.6 × 10²³

12. The number of molecules present in 8 g of oxygen gas is …………….
(A) 6.022 × 10²³
(B) 3.011 × 10²³
(C) 12.044 × 10²³
(D) 1.505 × 10²³
Answer:
(D) 1.505 × 10²³

13. The number of molecules in 22.4 cm³of ozone gas at STP is ……………….
(A) 6.022 × 10²⁰
(B) 6.022 × 10²³
(C) 22.4 × 10²⁰
(D) 22.4 × 10²³
Answer:
(A) 6.022 × 10²⁰

14. 11.2 cm³ of hydrogen gas at STP, contains …………….. moles.
(A) 0.0005
(B) 0.01
(C) 0.029
(D) 0.5
Answer:
(A) 0.0005

15. The mass of 224 mL of hydrogen gas at STP is
(A) 0.02 g
(B) 0.224 g
(C) 2.24 g
(D) 20.0 g
Answer:
(A) 0.02 g

16. 4.4 g of an unknown gas occupies 2.24 L of volume under STP conditions. The gas may be ………………
(A) CO₂
(B) CO
(C) O₂
(D) SO₂
Answer:
(A) CO₂