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Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 1.
What is a carbonyl group?
Answer:
Carbonyl group : A functional group in which a carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free is called a carbonyl group and represented as . Carbonyl group is present in aldehydes and ketones.

Question 2.
What are carbonyl compounds?
Answer:
The organic compounds containing a carbonyl group are called carbonyl compounds. For example, acetaldehyde, , acetone, . As carbonyl group is common in aldehydes and ketones, their methods of preparation and properties show similarities.

Question 3.
What are carboxylic compounds?
Answer:
The compounds in which the functional group is – COOH are known as carboxylic compounds. Due to the – OH group bonded to group, carboxylic acids are distinct from aldehydes and ketones.

Question 4.
How are carbonyl compounds classified ?
OR
Name the compounds containing carbonyl group.
Answer:
The carbonyl compounds contain a group . They are classified as follows :

Question 5.
What are aliphatic aldehydes?
Answer:
The compounds in which the – CHO group (formyl group or aldehyde group) is attached directly to sp³ hybridized carbon atom that is saturated carbon atom are called aliphatic aldehydes. (Exception : Formaldehyde, H – CHO is also classified as aliphatic aldehyde though – CHO group is not attached to any carbon).

For example :

Question 6.
What are aromatic aldehydes ?
Answer:
The compounds in which – CHO group is attached directly to an aromatic ring are called aromatic aldehydes.

For example :

Question 7.
Explain the structure of carbonyl functional group.
Answer:

• In the carbonyl functional group, carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free, and it is represented as .
  
• The carbonyl carbon atom is sp²-hybridised forming coplanar three sigma (σ) bonds with the bond angle 120°.
• One sigma bond is formed with oxygen atom while other two sigma (σ) bonds are formed with hydrogen or carbon atoms.
• The remaining unhybridised 2p_z orbital of carbon atom overlaps with p orbital of oxygen atom colaterally forming a pi (π) bond. Hence, carbon atom is joined to oxygen atom by a double bond of which one is sigma and another is n.
• The oxygen atom in the carbonyl group has two lone pairs of electrons.
• The carbonyl bond is strong, short and polarized.
• The polarity of the carbonyl group is explained on the basis of resonance involving neutral and dipolar structures as shown below :
  

Question 8.
What are aliphatic ketones? How are they classified?
Answer:
Aliphatic ketones : The compounds in which group is attached to two alkyl groups are called aliphatic ketones.

Ketones are classified into two types :

1. Simple or symmetrical ketones and
2. mixed or unsymmetrical ketones.

1. Simple or symmetrical ketone : The ketone in which the carbonyl carbon is attached to two identical alkyl groups is called a simple or symmetrical ketone.

2. Mixed or unsymmetrical ketone : The ketone in which the carbonyl carbon is attached to two different alkyl groups is called a mixed or unsymmetrical ketone.

Question 9.
What are aliphatic carboxylic acids? Give their general formula.
Answer:
The organic compounds in which carboxyl (- COOH) group is bonded to an alkyl group are called aliphatic carboxylic acids or fatty acids. (Exception : Formic acid, H-COOH is also classified as aliphatic carboxylic acid though-COOH group is not attached to any carbon).

For example :

Question 10.
How are carboxylic acids classified ? Give examples. (3 marks)
Answer:
Carboxylic acids are classified according to the presence of number of carboxyl groups into mono-, di-, tri- and polycarboxylic acids.

• Monocarboxylic acids : These carboxylic acids contain one carboxyl group.
  Example :
• Dicarboxylic acids : These contain two carboxyl groups
  Example :
• Tricarboxylic acid : These contain three carboxyl groups
  

Question 11.
What are aromatic carboxylic acids ? Give examples.
Answer:
Aromatic carboxylic acids : These are the compounds in which one or more carboxyl groups (- COOH) are attached directly to the aromatic ring.

For example :

Question 12.
Give examples of common carboxylic acids which are used in daily life.
Answer:
Common carboxylic acids are widely distributed in nature, they are found in both the plants and animals.

• Acetic acid is main constituent of vinegar.
• Butyric acid of butter which is responsible for odour of rancid butter.
• L-lactic acid is present in curd.
• Citric acid is found in citrus fruits.
• Higher carboxylic acids such as palmitic acid, stearic acid and oleic acid are the components of animal fats and vegetable oils.

Nomenclature of Aldehydes :

(A) Common System :

• The names of aldehydes are derived from the common names of acids.
• The suffix ‘-ic acid’ of an acid is replaced by ‘aldehyde’.
• The positions of the substituents in the molecule are indicated by Greek letters α, β, γ, etc.
• starting from the carbon atom attached to the carbonyl group. E.g.
  

(B) I UP AC System :

• The longest carbon atoms chain containing aldehyde carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘al’. Alkane → Alkanal
• The position (locant) of aldehyde group need not be mentioned since it is always at the end position.
• The substituents in the alkyl group are prefixed in an alphabetical order by appropriate locants.
• When two – CHO groups are present at the two ends of the chain the ending ‘e’ of alkane is retained and the suffix  ‘-dial’ is added to the name of parent aldehyde.
• In IUPAC nomenclature an alicyclic compound -in which – CHO group is attached directly to the ring is named as a carbaldeliyde. The suffix ‘carbaldehyde’ is added after the full name of parent cycloalkane structure.

(C) Common or trivial names :

(1) The common name of a carboxylic acid is derived from the source from which it was first isolated.
The following table gives common names and the source or origin of name.

(2) In branched carboxylic acids, the position of substituents are indicated by Greek alphabet.

For example :

(D) IUPAC system of nomenclature :

• The longest continuous cha in of carbon atoms including the carbon atom of – COOH group ¡s selected.
• The carboxvlic acid is conside red (IS a derivative of the corresponding parent a/kane.
• The carbon atom of the – COOH group is always at terminal position, hence need not to be indicated while writing IUPAC name.
• The position of the other substitutents are indicated by the appropriate locants in alphabetical order.
• In case of dicarboxylic acids, ‘dioic acid’ is added to parent alkane.
• In an alicyclic compound having a carboxyl group directly attached to alicyclic ring is named as cycloalkane carboxylic acid.

Trivial and IUPAC names of carboxylic acids and aldhydes

Question 13.
Give the common names and IUPAC names of the Miowing aldehydes :
Answer:

Question 14.
Write the structures of the following compounds :
Answer:

Question 15.
What is the IUPAC name of the following compound?

Answer:
IUPAC name : 2-Amino butanoic acid

Question 16.
Write IUPAC name of

Answer:
IUPAC name : Ethanedioic acid

Question 17.
Write the structure and give IUPAC names of following carboxylic acid.

Answer:

Question 18.
Draw the structures of the following compounds :
Answer:

Question 19.
Give IUPAC names of the following carboxylic acids.
Answer:

Question 20.
Write the structures and IUPAC names of all isomers of carboxylic acids having molecular formula C₅H₁₀O₂. HOW many of them are chiral?
Answer:

Nomenclature of ketones :

(A) Common System :

• Ketones are named according to the alkyl groups attached to the carbonyl carbon atom followed by the word ketone.
• The substituents in the alkyl groups are indicated by Greek letters a, f, y, etc. starting from the carbon atom attached to the carbonyl group.

(B) IUPAC System :

• The longest continuous chain containing carbonyl carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘one’. Alkane → Alkanone
• The position of carbonyl group is represented by the lowest locant.
• The substituents in the alkyl groups are prefixed in the alphabetical order along with their positions by appropriate locants.
• When two C = O groups are present, then ending ‘e ’ of alkane is retained and the suffix – ‘dione ’ is added to the name of parent ketone indicating the locants of ketonic carbonyl groups.
• In case of polyfunctional ketones, higher priority group is given lower number.
• When ketonic carbonyl is a lower priority group it is named as ‘oxo’, preceded by the locant. In alicyclic ketones, carbonyl carbon is numbered as 1.

Question 21.
Give the common and IUPAC names of the following ketones :
Answer:

Question 22.
Give the structures of the following compounds :
Answer:

Question 23.
Give IUPAC names of the following compounds :
Answer:

Question 24.
Write the structures and give common names and IUPAC names of the carbonyl compounds represented by formula C₅H₁₀O.
Answer:

Question 25.
Write the structure and give IUPAC names of the following compounds :
Answer:

Question 26.
Write the structure of the following compounds :
Answer:

Question 27.
How is an aldehyde obtained from an alcohol ?
Answer:
When a primary alcohol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, an aldehyde is obtained.

For example, when ethanol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, acetaldehyde (ethanal) is obtained.

Question 28.
How is ketone obtained from an alcohol?
Answer:
When a secondary alcohol is oxidized with potassium dichromate and dii. H₂SO₄ under controlled conditions, a ketone is obtained.

For example. when 2-propanol is oxidized with potassium dichromate and dii, H₂SO₄ under controlled conditions, accIone (propanone) is obtained.

Question 29.
How are the following compounds obtained from alcohol:
(1) Methanal
(2) Propanal
(3) BuLanal
(4) 3-Methylpentanal?
Answer:

1. Mehanol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms methanal.
   
2. Propan-1-ol on controlled oxidation with K,Cr20 and dilute H₂SO₄ forms propanal.
   
3. Butan..l-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanal.
   
4. 3-Mcthylpcntan.1-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ gives 3-Methylpentanal.
   

Question 30.
How are the following compounds prepared from alcohol :
(1) Butanone
(2) Pentan-3-one
(3) 2,2-Dimethylpropanal?
Answer:

1. Butan-2-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanone.
   
2. Pentan-3-ol on oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms Pentan-3-one.
   
3. 2,2-Dimethylpropan- I -ol on oxidation with K₂Cr₂O₇ or KMnO₄ and dilute H₂SO₄ forms 2,2-Dimethyipropanal.
   

Question 31.
How is an aldehyde obtained from primary alcohol ?
Answer:
When vapours of primary alcohol is passed over heated copper at 573 K, dehydrogenation takes place, an aldehyde is formed.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K, acetone (propanone) is obtained.

Question 32.
How is ketone obtained from secondary alcohol?
Answer:
When vapours of secondary alcohol is passed over heated copper at 573 K. dehydrogenation takes place, a ketone is obtained.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K. acetone (propanone) is obtained.

Question 33.
How are the following compounds obtained from alkene :
(1) Formaldehyde
(2) Acetaldehyde and
(3) Acetone ?
Answer:
When a stream of ozonised oxygen is passed through a solution of an alkene, in organic solvent, an unstable addition cyclocompound, ozonide is formed which on reduction with zinc dust and water forms an aldehyde or a ketone or a mixture of both.

1. Formaldehyde : Under these conditions ethylene gives formaldehyde.
   
2. AcetuIdeh&: Symmetrically disubstituted alkene like but-2-ene gives acetaldehyde.
   
3. scctnne: Tetrasubshtwed alkene like 2,3-dimethyl but-2-ene gives acetone.
   

Question 34.
Write ozonolysis reaction for
(1) Propylene and
(2) Isobutylene.
Answer:
(1) Propylene on reaction with ozonised oxygen in the organic solvent forms propylene ozonide which on reduction with zinc dust and water forms acetaldehyde and formaldehyde.

Question 35.
How are the following compounds obtained from alkynes :
(1) Acetaldehyde
(2) Acetone?
Answer:

1. Acetaldehyde : On passing acetylene through warm 40% H₂SO₄ in the presence of 1 % H_gSO₄, vinyl alcohol is obtained which tautomerises and forms acetaldehyde. It is a hydration reaction.
   
2. Acetone : On passing propyne through warm 40 % H₂SO₄ in the presence of 1 % H_gSO₄, alkenol is obtained which on tautomerisation form acetone.
   

Question 36.
Predict the products when
(1) dimethyl acetylene
(2) ethyl acetylene and
(3) diethyl acetylene are treated with mercuric sulphate in dilute sulphuric acid.
Answer:
(1) Dimethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms ethyl methyl ketone by tautomerisation.

(2) Ethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms Butan-2-one by tautomerisation.

(3) Diethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms hexan-3-one.

Question 37.
Write the structures of aldehydes and ketones obtained by ozonolysis.

Answer:

Answer:

Question 38.
Predict the structures of ketones produced by hydration of but-l-yne and but-2-yne.
Answer:

Question 39.
How is acetaldehyde prepared from acetyl chloride?
Answer:
Acetyl chloride is reduced to acetaldehyde by hydrogen in presence of Pd catalyst poisoned with BaSO₄. This reaction is called Rosenmund reduction.

Question 40.
How is benzaldehyde obtained from benzoyl chloride?
OR
Write chemical equation for Rosenmund reduction.
Answer:
When benzoyl chloride is hydrogenated in the presence palladium on barium sulphate (Pd/BaSO₄), benzaldehyde is obtained. This reaction is called Rosenmund reduction.

Question 41.
How will you prepare acetophenone from benzene? (Friedel – Crafts acylation).
Answer:
When benzene is treated with acetyl chloride in the prcscncc of anhydrous aluminium chloride, acetophenonc is obtained. This reaction is known as Friedel – Crafts acylation.

Question 42.
How will you convert benzene into 1-phenylethanone?
Answer:

Question 43.
How will you obtain 4-Nitrobenzaldehyde from 4-Nitrotoluene ? (Friedel- Crafts reaction).
Answer:
When 4-nitrotoluene is treated with chromium oxide in acetic anhydride, a diacetate derivative is obtained which on acid hydrolysis produces 4-nitrobenzaldehyde.

Question 44.
How will you prepare Propanone (acetone) from Grignard reagent?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When acetyl chloride reacts with dimethyl cadmium, propanone (acetone) is obtained.

Question 45.
How is acetophenone obtained from Grignard reagent ?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When benzoyl chloride reacts with dimethyl cadmium, acetophenone is obtained.

Question 46.
How is benzyl methyl ketone obtained from Grignard reagent ?
OR
Convert: Acetyl chloride to benzyl methyl ketone.
Answer:
Grignard reagent (Benzyl magnesium chloride) reacts with cadmium chloride to give diphenyl cadmium. When acetyl chloride reacts with dibenzyl cadmium, benzyl methyl ketone is obtained.

Question 47.
How is an aldehyde obtained from alkyl nitrile ?
OR
What is Stephen reaction ?
OR
Write a note on Stephen reaction.
Answer:
(1) An ethereal solution of a nitrile is reduced to imine hydrochloride by SnCl₂ in the presence of HCl gas. Further, imine hydrochloride on acid hydrolysis gives aldehyde. This reaction is called Stephen reaction.


(2) Alternatively, nitriles are selectively reduced by diisobi.ityl aluminium hydride (DIBAI-H) lo imines which on acid hydrolysis to aldehydes.

Question 48.
How are following compounds prepared using Gngnard reagent
(1) Acetone
(2) Benzophenone?
Answer:
(1) Acetone: Acetoniti-ile (ethanenitrile) reacts with methyl magnesium iodide in presence of dry ether to give imine complex which on hydrolysis gives acctonc. During reaction acetonitrile and methyl magnesium iodide should be
taken in equimolecular proportion.

(2) Benzophenone: Benzonitrile reacts with phenyl magnesium bromide in presence of dry ether to give an imine complex which on acid hydrolysis gives a benzophenone. During reaction bcnzonitrile and phenyl magnesium bromide should be aken in equimolecular proportion.

Question 49.
Write the structures and IUPAC names of ketones produced by Friedel-Crafts acylation of benzene with
(i) C₂H₅COCl
(ii) C₆H₅COCl.
Answer:

Question 50.
Predict the products of the following conversions.
Answer:

Question 51.
How are the following preparations carried out ?

(1) Benzaldehyde from toluene. (Etard oxidation)
Answer:
When toluene is treated with solution of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde. This reaction is called Etard reaction.

(2) Benzaldehyde from methyl arene.
Answer:
Methylarene is converted into a benzyllidene diacetate on treatment with chromium oxide in acetic anhydride at 273-278 K. The diacetate derivative on acid hydrolysis gives benzaldehyde.

(3) Benzaldehyde from toluene (commerical method).
Answer:
Side chain chlorination of toluene gives benzal chloride which on hydrolysis gives benzaldehyde.

(4) Benzaldehyde from benzene (Gattermann-Koch synthesis).
OR
Write chemical equation for Gatter- mann-Koch formylation.
Answer:
When benzene is treated with vapours of carbon monoxide and hydrogen chloride in the presence of a catalyst mixture of A1Cl₃ and CuCl under high pressure, benzaldehyde is obtained. This reaction is called Gattermann- Koch synthesis.

Preparation of aromatic ketones from hydrocarbons

Question 52.
Explain Friedel-Craft’s acylation reaction.
Answer:
The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 53.
Give the preparation of acetophenone from benzene using
(i) acetyl chloride
(ii) acetic anhydride.
Answer:

The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 54.
How will you prepare propionaldehyde from ethyl propionate?
Answer:
When ethyl propionate is reduced in presence of diisobutyl aluminium hydride (DIBAI-H), propionaldehyde is obtained.

Question 55.
Explain the structure of carboxyl group.
Answer:
In carboxyl group, the carboxyl carbon is sp²-hybridised and the bonds to the carboxyl carbon lie in one plane. The C-C = O and O = C-O bond angles are 120°. The carboxylic carbon is less electrophilic than carbonyl carbon because of the resonance structures shown below :

Question 56.
How is carboxylic acid obtained by the acid hydrolysis of an alkyl cyanide ?
Answer:
Alkyl cyanides or alkyl nitriles on acid or alkaline hydrolysis give corresponding carboxylic acids.

Acid Hydrolysis of Alkyl ankle: When alkyl cyanide is boiled wiLh dilute mineral acid, it gives corresponding carboxylic acid. In this, acid amide is obtained as the intermediate product.

Question 57.
How is ethanoic add obtained from methyl cyanide by acid hydrolysis?
Answer:
When methyl cyanide is heated with dilutc hydrochloric acid or dilute sulphuric acid. ethanoic acid is obtained.

Question 58.
How Is proplonlc acid obtained from an alkyl nitrile?
Answer:
When ethyl cyanide (propionitrile) is boiled with dilute HCI or dilute H₂SO₄, propionic acid is obtained.

Question 59.
How Is benzoic acid obtained from bcnzamide?
Answer:
When benzarnide is heated with dil. HCl. benzoic acid is obtained.

Question 60.
How is carboxylic acid obtained from acyl chlorides and acid anhydrides?
Answer:
When acyl chloride is hydrolysed with water, carboxylic acid is obtained. The reaction is carried out in presence of a base pyridine or NaOH to remove HCl generated.

Acetyl chloride reacts with water almost explosively while benzoyl chloride very slowly.

Acid anhydrides react with water to give carboxylic acid.

Question 61.
How is benzoic acid obtained from
(i) ethyl benzoate
(ii) styrene?
Answer:
(i) Benzoic acid from ethyl benzoate : When an ethyl benzoate is heated with dil. H₂SO₄, undergoes hydrolysis to form benzoic acid and ethyl alcohol.

(ii) Benzoic acid from styrene:

Question 62.
How is propanoic acid obtained from phenyl propanoate?
Answer:
When phenyl propanoate is heated with dil. NaOH, sodium salt is obtained, which on hydrolysis gives propanoic- acid.

Question 63.
How is propanoic acid obtained from methyl propanoate ?
OR
When methyl propanoate is heated with dil. NaOH, sodium, salt is obtained, which on hydrolysis gives propanoic acid.
Answer:

Question 64.
How is benzoic acid obtained from phenyl ethene?
Answer:
When phenyl ethene is heated with strong oxidising agents like acidic KMnO₄ or acidic K₂Cr₂O₇, benzoic acid is obtained.

Question 65.
How is adipic acid obtained from cyclohexene?
Answer:
When cyclohexene is heated with acidified KMnO₄, adipic acid (Hexane-1, 6-dioic acid) is obtained.

Question 66.
What is carbonation of Grignard reagent ? How is acetic acid prepared by this reaction ? How is ethanoic acid pepared from dry ice?
Answer:
Addition reaction of carbon dioxide (0 = C = 0) to Grignard reagent, forming a complex and further formation of carboxylic acid is called carbonation of Grignard reagent.

Example : When methyl magnesium iodide is added to solid carbon dioxide, a complex is formed which on acid hydrolysis forms acetic acid.

Question 67.
How is benzoic acid prepared from Grignard reagent?
OR
Write the preparation of benzoic acid from dry ice.
Answer:
When phenyl magnesium bromide is treated with dry ice (solid carbon dioxide) in the presence dry ether, complex is obtained which on acidification gives benzoic acid.

Question 68.
What are soaps ? How are they prepared ?
Answer:
The sodium or potassium salts of higher fatty acids are known as soaps. Soaps contain more than twelve carbon atoms.

When fat or oil is hydrolysed using sodium or potassium hydroxide solution, soap obtained remains in colloidal form. Soap and glycerol are separated by adding sodium chloride. Soap precipitates out due to common ion effect, and glycerol remains in the solution can be recovered by fractional distillation.

Question 69.
How is benzoic acid prepared from alkyl benzenes ?
OR
How will you convert the following :
(1) n-butyl benzene to benzoic acid.
(2) Toluene to benzoic acid.
(3) Cumene to benzoic acid ?
Answer:
When an alkyl benzene is heated with strong oxidizing agents like acidic or alkaline KMnO₄ or acidified K₂Cr₂O₇ etc. gives aromatic carboxylic acid. The alkyl side chain gets oxidised to -COOH group irrespective of the size of the chain.

Question 70.
Lower aldehydes and ketones are water soluble whereas higher homologues are insoluble. Explain, why.
Answer:
(1) The oxygen atom of shows hydrogen bonding with water molecule.

(2) As a result of this, the lower aldehydes and ketones are water soluble (example – acetaldehyde, acetone). As the molecular mass increases, the proportion of hydrocarbon part of the molecule increases which cannot form hydrogen bonding with water and the solubility of higher homologues in water decreases.

Question 71.
Carboxylic acids have higher boiling points than those of alcohols, aldehydes, ketones, ethers, hydrocarbons of comparable molecular masses. Explain, why.
Answer:
(1) Carboxylic group (-COOH) in acids is highly polar. in liquid state, pair of carboxylic acid molecules is held together by two intermolecular hydrogen bonds, have higher aggregations and in the vapour.

(2) Intermolecular hydrogen bonding in carboxylic acids state most of the carboxylic acid.s exist as dimmers in which two molecules are held by two hydrogen R – R bonds. Acidic hydrogen of one molecule forms hydrogen bond with carbonyl oxygen of the other molecule. This doubles the size of the molecule resulting in increase in o – intermolecular van der Waals forces, which in turn results in high boiling points. Therefore, carboxylic acids possess higher boiling points than those of alcohols, aldehydes, ketones. ether, hydrocarbons of comparable molecular masses.

Question 72.
Lower aliphatic carboxylic acids are miscible with water while higher carboxylic acids are immiscible.
Answer:
(1) Lower aliphatic carboxylic acids are miscible with water due to the formation of hydrogen bonding with water molecules.

(2) Hydrogen bonding between acid and water. As the molecular mass increases (he solubility of carboxylic acids in water decreases. The insolubility of carboxylic acids is due to increased hydrophobic interaction of hydrocarbon parts with water.

Question 73.
Explain why carboxylic acids are much weaker acids than mineral acids.
Answer:
Carboxylic acids are the organic compounds which are acidic in nature. However, compared to mineral acids like HCI or H₂SO₄. the carboxylic acids are weaker acids.

The strength of acidity depends upon their ability to release H⁺ ions. Greater the ease with which they release H⁺ ions, stronger is the acid.

Carboxylic acids when dissolved in water, pcoduce H⁺ due to its dissociation. (it does not dissociate completely.)

The mineral acid-s like HCI, H₂SO₄ release H⁺ ion to a larger extent as they dissociatc almost complctcly in aqueous solution for e.g. HCl → H⁺ Cl^– thus carboxylic acids are weaker than mineral acids. The equilibrium exists in aqueous solution of carboxylic acid as

Since concentration of water practically remains constant

where K_a is acidity constant
Larger the value K_a greater is the extent of ionization and stronger is the acid. But strength of acids is expressed in ternis of their pK_a values. Smaller the value of pK_a. the stronger is the carboxylic acid. Here pK_a value of carboxylic acids is higher than mineral acids. Hence, carboxylic acids arc weaker than mineral acids.

Question 74.
Carboxylic acids are more acidic than phenols and alcohols. Explain. Why?
Answer:
(1) Carboxylic acid loses a proton as compared to phenol. Consider the ionization of carboxylic acid and phenol

Due to delocalization the negative charge over the ortho and para positions of aromatic ring, phenoxide anion is more stable than phenol. Thus phenol easily undergoes ionization

However, alcohol and alkoxide ion are single structures. In an alkoxide anion the negative charge is localized on a single oxygen atom. Hence, phenols are more acidic than alcohols.

(2) Carboxylic acid has two resonance hybride non equivalent structures (I & II) while carboxylate anion has two resonance hybrid equivalent structures (III & IV). The carboxylate ion is more stable than carboxylic acid and equilibrium is shifted towards the direction of increased ionization.

(3) Carboxylate ion has two equivalent resonance structures with nejative charge is delocalized over two electro negative oxygen atoms. Phenoxide anion has non-equivalent resonance structures in which negative charge is
delocalized over one oxygen atom and less electronegative carbon atom. As a result carboxylate anion is more stable than phenoxide ion. Hence carboxylic acids ionize to the greater extent than phenol furnishing higher concentration of H⁺ ions. Therefore, carboxylic acids are more acidic than phenols and alcohols.

Question 75.
Trichloro acetic acid is a stronger acid than acetic acid. ExplaIn.
Answer:
(1) The acidic nature of carboxylic acid is due to the ability to release H ions. Greater the ease with which they release H⁺ ions, stronger is the acid. Any factor that stabilizes the carboxylate ion would help the release of H⁺
ions and thus increase the strength of the acid. The electron-withdrawing group attached to -carbon atom increases the strength of the acid. In trichloroacetic acid, three chloro substituents on s-carbon atom of acetic acid makes the electrons withdrawing effect more pronounced and the negative charge of carboxylate ion formed gets dispersed.

Thus, increases the stability of carboxylate ion.

(2) The acetate ion formed gets destabilised due to the electron releasing effect of a methyl group (+ I effect). As a result, acetic acid dissociates to a lesser extent.

(3) The trichloro acetate ion formed gels stabilised due to electron-withdrawing effect of three 3Cl atoms (- I effect). As a result. trichloro acetic acid dissociates to a greater extent. Trichioro acetic acid having lower pK_a value than acetic acid. Hence. trichloro acetic acid is a stronger acid than acetic acid.

Question 76.
Which is the stronger acid in each of the following pairs ?
(1) CH₃-COOH and CH₂ = CH-COOH
Answer:
CH₂ = CH-COOH is the stronger acid than CH₃-COOH

(2) C₆H₅-COOH and C₆H₅-CH₂-COOH
Answer:
C₆H₅-COOH is the stronger acid than C₆H₅-CH₂-COOH

Answer:

(4) CH₃-CH₂-COOH and NC-CH₂-COOH
Answer:
NC-CH₂-COOH is the stronger acid than CH₃-CH₂-COOH

(5) (CH₂)₂CH-CH₂-COOH and (CH₃)₂NH-CH₂-COOH
Answer:
(CH₃)₂NH-CH₂-COOH is the stronger acid than (CH₃)₂CH-CH₂-COOH

(6) O₂N-CH₂-C00H and Cl-CH₂-COOH
Answer:
NO₂-CH₂-COOH is the stronger acid than Cl-CH₂-COOH.

Question 77.
Arrange the following acids in their increasing order of acidic strength.

(1) Acetic acid, chloroacetic acid, propionic acid, formic acid.
Answer:
Propionic acid < acetic acid < formic acid < chloroacetic acid

(2) Bromoacetic acid, chloroacetic acid, fluoroacetic acid, iodoacetic acid.
Answer:
Iodoacetic acid < bromoacetic acid < chloroacetic acid < fluoroacetic acid.

(3) Acetic acid, chloroacetic acid, dichloroacetic acid, trichloroacetic acid.
Answer:
Acetic acid < chloroacetic acid < dichloroacetic acid < trichloroacetic acid.

(4) n-butyric acid, 3-chlorobutyric acid, 2-chlorobutyric acid, 3-chlorobutyric acid.
Answer:
n-Butyric acid < 3-chlorobutyric acid < 2-chlorobutyric acid < 1-chlorobutyric acid.

(5) Acetic acid, benzoic acid, p-methoxy benzoic acid, p-nitrobenzoic acid.
Answer:
Acetic acid < benzoic acid < p-methoxy benzoic acid < p-nitrobenzoic acid.

(6) Acetic acid, phenyl acetic acid, p-nitro phenyl acetic acid.
Answer:
Acetic acid < phenyl acetic acid < p-nitro phenyl acetic acid.

(7) Benzoic acid, p-toluic acid, p-chlorobenzoic acid.
Answer:
p-toluic acid < benzoic acid < p-chlorobenzoic acid.

Question 78.
Arrange the following carboxylic acids in order of increasing acidity : m-Nitrobenzoic acid, Trichloroacetic acid, benzoic acid, a-Chlorobutyric acid.
Answer:
Acidity in the increasing order : Benzoic acid, m-nitrobenzoic acid, a-chlorobutyric acid, trichloroacetic acid.

Question 79.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Question 80.
Explain polarity of carbonyl group.
Answer:
The polarity of a carbonyl group is duc to higher electronegativity of oxygen compared to carbon. It is explained on the basis of resonance involving neutral and dipolar structures.

The carbonyl carbon has positive polarity (see structures (A) and (D)). Therefore, it is electron deficient. As a result, this carbon atom is electrophilic (electron loving) and is susceptible to attack by a nucleophile (Nu : ^–).

Question 81.
Explain SchifPs reagent test.
OR
What is a SchifPs reagent ? How is it used to detect aldehydes ?
Answer:

• Schiff’s reagent is prepared by dissolving pink p-rosaniline hydrochloride (dye Fuchsin) in water and passing SO₂ gas till the pink solution is decolourised.
• Schiff s reagent is an oxidising agent.
• When an aldehyde is added to Schiff s reagent, the colourless solution turns pink or in magenta colour and aldehyde is oxidised to a carboxylic acid.
• This test is not given by ketones, hence, used to distinguish between aldehyde and ketone.

Question 82.
Which colour is obtained when SchifFs reagent is treated with acetaldehyde?
Answer:
When Schiff s reagent is treated with acetaldehyde, pink colour is obtained.

Question 83.
What is Tollen’s reagent?
Answer:

• Tollen’s reagent is an ammoniacal silver nitrate, [Ag(NH₃)₂]⁺ OH^–.
• It is prepared by adding NH₄OH solution to silver nitrate solution.
  
• It is a stronger oxidising agent than Fehling solution. Aldehyde when heated with Tollen’s reagent, silver mirror is deposited.

Question 84.
Explain Tollen’s reagent test.
OR
Explain silver mirror test.
Answer:

• Tollen’s reagent is an (ammoniacal silver nitrate) [Ag(NH₃)₂]⁺ OH^–.
• When an aldehyde, like acetaldehyde is heated with Tollen’s reagent, it is oxidised to acetic acid and silver ions Ag⁺ in Tollen’s reagent complex are reduced to silver Ag giving greyish black precipitate or deposition of silver on inner surface of the test tube which shines like a mirror. Hence this test is also called silver mirror test.
  
• This test is not given by ketones.
• Hence Tollen’s reagent is used to distinguish between aldehydes and ketones.

Question 85.
What is Fehling’s solution? How is it prepared?
Answer:

• Fehling’s solution is a complex of cupric ions with tartaric acid.
• It is a mild oxidising agent.
• It is prepared by mixing equal amount of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochelle salt) in caustic soda (NaOH) solution.
• It is used to detect aldehydes that decolourise deep blue colour of the solution and give red precipitate of Cu₂O.

Question 86.
Explain Fehling’s solution test.
Answer:

• Fehling’s solution is a mixture of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochella salt) in caustic soda (NaOH) solution.
• When an aldehyde is heated with Fehling’s solution, the deep blue colour of the solution disappears and Cu⁺² (cupric ion) is reduced to Cu⁺ ion a red precipitate of cuprous oxide, Cu₂O is obtained while aldehyde is oxidised to a carboxylate ion.
• For example,
  
• This test is not given by ketones, since they cannot be oxidised by Fehling solution.
• Aromatic aldehydes are not oxidised by Fehling solution.
• Hence this test is used to distinguish between aldehydes and ketones.

Question 87.
What is the action of the following reagents on ethanal :
(1) Fehling’s solution,
(2) Tollen’s reagent or Ammonical silver nitrate ?
Answer:

1. When ethanal is heated with Fehling’s solution, the deep blue colour of the solution disappears and a red precipitate of Cu₂O is obtained.
   
2. When ethanal is heated with Tollen’s reagent a greyish black precipitate or deposition of silver is obtained.
   

Question 88.
Why is benzaldehyde NOT oxidized by Fehling solution ?
Answer:
When benzaldehyde is treated with Fehling solution, it does not reduce cupric ion (Cu⁺²). Fehling solution does not oxidise benzaldehyde. Thus, Fehling test cannot be used for aromatic aldehyde.

Question 89.
Explain laboratory test for ketonic group or sodium nitroprusside test.
Answer:
Laboratory test for ketonic group : Sodium nitroprusside test : When a freshly prepared sodium nitroprusside solution is added to a ketone, mixture is shaken well and basified by adding sodium hydroxide solution drop by drop, red colour appears in the solution, which indicates the presence of ketonic (> C = O) group.

The anion of ketone formed by alkali reacts with nitroprusside ion to form a red coloured complex which indicates the presence of the ketonic group.

Question 90.
What is the action of hydrogen cyanide on the following :
(1) Acetaldehyde
(2) Acetone
(3) Benzaldehyde?
Answer:
(1) Action of HCN on acetaldehyde : When acetaldehyde is treated with hydrogen cyanide, acetaldehyde cyanohydrin is formed.

(2) Action of HCN on acetone : When acetone is treated with hydrogen cyanide, acetone cyanohydrin is formed.

(3) Action of HCN on benzaldehyde : When benzaldehyde is treated with hydrogen cyanide, benzaldehyde cyanohydrin is formed.

Question 91.
What is the action of hydrogen cyanide in basic medium on (1) butanone (2) 2,4-dichlorobenzaldehyde?
Answer:
(1) Action of hydrogen cyanide on butanone : When butanone is treated with hydrogen cyanide, butanone cyanohydrin is obtained.

(2) Action of hydrogen cyanide on 2,4-dichlorobenzaldehyde : When 2, 4-dichloro benzaldehyde is treated with hydrogen cyanide, cyanohydrin of 2,4-dichloro benzaldehyde is obtained.

Question 92.
What is the action of sodium bisulphite on :
(1) Acetaldehyde
(2) Acetone (propanone)?
Answer:
(1) Acetaldehyde reacts with saturated aqueous solution of sodium bisulphite (NaHSO₃) and forms crystalline acetaldehyde sodium bisulphite. It is water soluble crystalline solid.

(2) Acetone reacts with saturated aqueous solution of sodium bisuiphite (NaHSO₃) and forms crystalline acetone sodium bisuiphite. It is water soluble crystalline solid.

Question 93.
A carbonyl compound ‘A’ having molecular formula C₅H₁₀O forms crystalline precipitate with sodium bisulphite and gives positive iodoform test but does not reduce Fehling solution. Write the structure of carbonyl compound.
Answer:
A carbonyl compound C₅H₁₀O has two structures.

Pentan-2-one forms crystalline precipitate with sodium bisulphite and gives positive iodoform test. But does not reduce Fehling solution.

Pentan-3-one does not react with iodine and NaOH because it does not contain group.

Question 94.
How does alcohols react with aldehydes and ketones?
Answer:
Aldehyde reacts with one molecule of anhydrous monohydric alcohol in presence of dry hydrogen chloride to give alkoxyalcohol known as hemiacetal, which further reacts with one more molecule of anhydrous monohydric alcohol to give a geminaldialkoxy compound known as acetal as shown in the reaction.

Ketones react with 1, 2 – or 1, 3 – diols in presence of dry hydrogen chloride to give five or six-membered cyclic ketals.

Question 95.
What is the action of ethanol on acetaldehyde ? What is the action of ethylene glycol on acetone ?
Answer:
Acetaldehyde reacts with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to form an intermediate known as hemiacetal, which further adds another molecule of alcohol to form a gem-dialkoxy compound known as acetal. Acetone reacts with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals.

Question 96.
Write the structure of product in the following reactions:

Answer:

Answer:

Question 97.
How does Grignard reagent react with the carbonyl compounds (or aldehydes and ketones)?
Answer:
The carbonyl compounds like aldehydes and ketones react with Grignard reagent (R – Mg – X) in dry ether and form a complex which on further hydrolysis with acid forms the corresponding alcohol.

Question 98.
What is the action of Grignard reagent, CH₃ – Mg – I on : (1) formaldehyde (2) acetone?
Answer:
(1) Grignard reagent with formaldehyde gives a primary alcohol.
Formaldehyde on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HCl forms ethyl alcohol.

(2) Acetone on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HC1 forms tert-butyl alcohol.

Question 99.
Explain the mechanism of addition reactions of ammonia derivatives H₂N-Z with carbonyl compounds (aldehydes or ketones).
Answer:
Derivatives of ammonia H₂N-Z reacts with carbonyl compounds (aldehydes or ketones) in weakly acidic medium to give addition products, which loses a water molecule to give a final product imine derivatives. A substituted imine is called a Schiff base. Schiff bases are solids and have sharp melting points.

General reaction :

Question 100.
What Is the action of ethylamine on :
(1) acetaldehyde
(2) acetone ?
Answer:
(1) Acetaldehyde on reaction with ethyl amine forms imine (Schiff base).

(2) Acetone on reaction with ethyl amine forms imine (Schiff base).

Question 101.
What are oximes? Which functional group do they contain?
Answer:
Oximes : These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with hydroxyl amine NH₂OH.

Question 102.
What is the action of hydroxyl amine (NH₂OH) on (1) acetaldehyde (2) acetone?
Answer:
(1) Acetaldehyde on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetaldoxime.

(2) Acetone on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetoxime.

Question 103.
What are hydrazones?
Answer:
Carbonyl compounds like aldehydes and ketones react with hydrazine forming compounds like hydrazones. For example, acetaldehyde on reaction with hydrazine gives acetaldehyde hydrazone.

Question 104.
Which compound can convert
Answer:
The compound which can convert.

Question 105.
What is the action of hydrazine on (1) formaldehyde (2) acetone ?
Answer:
(1) Formaldehyde on reaction with hydrazine forms formaldehyde hydrazone.

(2) Acetone with hydrazine forms acetone hydrazone.

Question 106.
What are phenylhydrazones?
Answer:
The carbonyl compounds like aldehydes and ketones on reaction with phenylhydrazine form hydrazones. For example, acetaldehyde on reaction with phenylhydrazine forms acetaldehydephenylhydrazone.

Question 107.
What is the action of phenylhydrazine on (1) formaldehyde (2) acetone (propanone) ?
Answer:
(1) Formaldehyde on reaction with phenylhydrazine forms formaldehydephenylhydrazone.

(2) Acetone on reaction with phenylhydrazine forms acetone phenylhydrazone.

Question 108.
What is the action of 2,4-Dinitrophenylhydrazine on
(1) Acetaldehyde
(2) Acetone
(3) Butanone
(4) Benzaldehyde ?
OR
Complete and rewrite the balanced chemical equation : Butanone + 2, 4-Dinitrophenylhydrazine.
Answer:
(1) Acetaldehyde on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(2) Acetone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(3) Butanone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

Question 109.
What is the action of semi carbazide on (1) Acetaldehyde (2) Acetone?
Ans.
(1) Acetaldehyde on reaction with semicarbazide forms scrnicarbazone derivative.

(2) Acetone on reaction with sernicarbazide forms semicarbazone derivative.

Question 120.
Write the structures of carbonyl compounds and ammonia derivatives that combine to give following imines.
Answer:

Question 121.
Write the structure of the products obtained from the following ketones by action of hydrazine in presence of (1) slightly acidic medium (2) strong base KOH.

Answer:
(1) In slightly acidic medium

(2) In the presence of a strong base KOH

Question 122.
Explain haloform reaction.
Answer:
A ketone containing -COCH₃ group is oxidised by sodium hypohalite a mixture of (sodium hydroxide and halogen) results in the formation of sodium salt of carboxylic acid having one carbon atom less than that of ketone and methyl group is converted to haloform.

Acetaldehyde is the only aldehyde which gives haloform reaction. In this reaction, R may be hydrogen, methyl group or aryl group and X may be Cl, Br or I. The reaction is given by all methyl ketones (CH₃ – CO – R) and all alcohols containing CH₃ – (CHOH) group.

When a methyl ketone is warmed with iodine and sodium hydroxide, a yellow precipitate of iodoform is obtained. The iodoform reaction is used as a qualitative test for detection of CH₃CO-group in a organic compound.

Question 128.
Identify the compounds, amongst the following, that give positive iodoform test.
(CH₃)₂CHOH, (CH₃)₃COH, CH₃COCH₂CH₂CH₃, CH₃CH₂CHO, CH₃CH₂CH(OH)CH₂CH₃, CH₃CH₂OH, C₆H₅COCH₂CH₃, CH₃CHO, C₆H₅CH₂CH₂OH and CH₃CH(OH)CH₂CH₂CH₃.
Answer:
For an iodoform test, the carbonyl compound must have  group.
The compounds that give positive iodoform test :

Question 129.
Explain cross aldol condensation.
Answer:
(1) An aldol condensation between two different carbonyl compounds (aldehydes and or ketones) takes place even though one of the two carbonyl compounds molecules does not contain a-hydrogen atom e.g. HCHO and C₆H₅CHO.

(2) If both aldehydes or ketones contain two a-hydrogen atoms each, then a mixture of four products, is formed. For example, a mixture of ethanal and propanal on reaction with dilute alkali followed by heating gives a mixture of four products.

Self aldol condensation:

Cross aldol condensation:

Question 130.
Write the structure of the major product of the following crossed aldol condensation.
Answer:
(1) Formaldehyde and propionaldehyde :

(2) Benzaldehyde with acetone:

Question 131.
Explain aldol condensation reaction of propionaldehyde.
Answer:
Since propionaldehyde has an a-hydrogen atom it undergoes aldol condensation with alkali Ba(OH)₂, forming 3-Hydroxy-2-methylpentanal.

3-Hydroxy-2-methylpentanal on heating undergoes dehydration and forms 2-Methylpent-2-enal.

Question 132.
If a mixture of formaldehyde and acetaldehyde is subjected to aldol condensation, predict the products formed and draw their structures.
Answer:
Since formaldehyde, does not have α-hydrogen atom it will not undergo self aldol condensation. Since acetaldehydehas a-hydrogen atom, it will undergo self aldol condensation.

Formaldehyde and acetaldehyde undergo cross aldol condensation.

Hence two aldol condensation products will be obtained.

Question 133.
Indicate by equations, what happens when a mixture of acetaldehyde and acetone are treated with alkali.
Answer:
When a mixture of acetaldehyde and acetone is treated with alkali, Ba(OH)₂, they undergo self aldol condensation and cross aldol condensation.
(1) Self aldol condensation acetaldehyde :

(2) Self aldol condensation of acetone:

(3) Cro5s aldol condensation:

Question 134.
Explain Cannizzaro reaction.
OR
Write a note on Cannizzaro reaction.
OR
Write a note on self oxidation-reduction reaction of aldehyde with suitable example.
Answer:

• Aldehydes which do not have a-hydrogen atom, on heating with concentrated alkali (50% aqueous or ethanolic solution of NaOH or KOH) undergo self oxidation and reduction reaction or redox reaction.
• This self redox reaction or disproportionation reaction is called Cannizzaro reaction.
• In this reaction one molecule of the aldehyde is oxidised to carboxylic acid while the second molecule of the aldehyde is reduced to alcohol (carboxylic acid formed, reacts with alkali, NaOH and forms a salt R – COONa).
• When formaldehyde (methanal) is heated with 50% NaOH solution, methanol (reduction product) and sodium formate (oxidation product) are formed.
  
• Ketones and aldehydes like acetaldehyde, propionaldehyde, etc. having a – H atom do not give Cannizzaro reaction.

Question 135.
Explain cross Cannizzaro reaction with example.
OR
Write the reaction for the action of 50 % NaOH on a mixture of formaldehyde and benzaldehyde.
Answer:
The reaction between two different aldehydes, not having a-hydrogen atoms is called cross Cannizzaro reaction. These two aldehydes undergo disproportionation in presence of concentrated alkali to give four products. However, if one of the aldehydes is formaldehyde, the reaction yields exclusively formate and alcohol to corresponding aldehyde.

Formaldehyde and benzaldehyde since do not have a-hydrogen atom, will undergo Cannizzaro (redox) reactions.

(1) Self Cannizzaro (redox) reaction :

(2) Cmss Cannizzaro (redox) reaction:

Question 136.
What is the action of cone, potassium hydroxide on benzaldehyde?
Answer:
When benzaldehyde is heated with concentrated potassium hydroxide in presence of methanol, a mixture of potassium benzoate and phenyl methanol (benzyl alcohol) is obtained.

Question 137.
Differentiate between Cannizzaro reaction and Aldol reaction.
Answer:

Cannizzaro reaction	Aldol reaction
1. It is given by aldehydes not having alpha hydrogen atom. 2.  In this reaction an aldehyde is converted to the corres­ponding acid and an alcohol. 3.  It is a disproportionate ion reaction. 4.  It requires concentrated alkali as a catalyst.	1. It is given by aldehydes and ketones possessing alpha hydrogen atom. 2.  In this reaction aldehydes and ketones are converted into aldol and ketols, respectively. 3.  It is an addition reaction. 4.  It requires dilute alkali as a catalyst.

Question 138.
Write the chemical equations for aldol condensation or Cannizzaro reaction that the following compounds undergo :
(1) Propanal
(2) 2-Methyl propanal (isobutyraldehyde)
(3) Pentanal
(4) 3-Methylbutanal
(5) Acetophenone
(6) p-Methoxybenzaldehyde
(7) 2-Methyl cyclohexanone
(8) Chloral
(9) Cyclopentanone
(10) Phenyl acetaldehyde
(11) 1-Phenyl propan-l-one.
Answer:
(1) Propanal (Aldol condensation) : Propanal contains α-H atom. Two molecules of propanal undergo self condensation in presence of dil. alkali to form 3-Hydroxy-2-methyl pentanal.

(2) 2-NIethvl propanal (Canniuaro reaction) : Two molecules of them undergo cannizzaro reaction in the presence of 50% alkali to form sodium isobutyrate and isohutyl alcohol.

(3) Pentanal (Aldol condensation) : Pentanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-propyl heptanal.

(4) 3-Methyl butanal (Aldol condensation) : 3-Methyl butanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-isopropyl-5-methyl hexanol.

(5) Acetophenone (Aldol condensation) : Acetophenone contains a-H atom. Two molecules of them undergo self condensation in the presence of base to form 3-Hydroxy-1, 3-diphenyl but-l-one.

(6) p-Methoxybenzaldehyde (Cannizzaro reaction) :

(7) 2-Methyl cyclohexanone

(8) Chloral (Cannizzaro’s reaction) : There is no α-H atom CCl₃CHO, therefore it undergoes Cannizzaro’s reaction.

(9) Cyclopentanone (Aldol condensation) : Cyclopentanone contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 2-(l-Hydroxy-1 cyclopentyl) cyclo pentane-l-one.

(10) Phenyl acetaldehyde (Aldol condensation) : Phenyl acetaldehyde contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2, 4-diphenylbutanal.

(11) 1-phenyl propan-l-one (Aldol condensation) : 1-phenyl propan-l-one contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2-methyl-l, 3-diphenyl pentan-l-one
3-Hydroy-2-methyl-1, 3-diphenyl pentan-l-one

Question 139.
Write a note on Clemmensen reduction.
OR
Explain Clemmeusen’s reduction.
OR
Explain the reduction of carbonyl group into methylene group.
Answer:

• The carbonyl groupon reduciion with zinc amalgam (Zn – Hg) in concentrated hydrochloric acid is converted into methylene group ( – CH₂ -).
  
• Aldehydes and kctoncs on reaction with Zn – Hg in concentrated HCl forms corresponding alkanes. ibis reduction is called Clemmensen reduction.
  
• Acetaldehyde on reduction with Zn – Hg in concentrated HCl forms ethane.
  
• Acetone on reduction with Zn – Hg in concentrated HCl forms propane.
  

Question 140.
Explain Wolff-Kishner reduction.
Answer:
Hydrazine (NH₂-NH₂) reduces carbonyl groupof aldehydes or ketones to metylene group  When aldehyde or ketone is heated with hydrazine in the presence of base such as potassium hydroxide and ethylene glycol, an alkane is obtained due to reduction of carbonyl compound.

Question 141.
Compound A (C₅H₁₀O) form a phenyl hydrazone and gives a negative Tollen’s reagent test and iodoform test. On reduction with Zn-Hg/HCl, compound A gives n-pentane. Write the structure of ‘A’.
Answer:
Since A (C₅H₁₀O) forms a phenyl hydrozone, it is a carbonyl compound. Since it gives negative Tollen’s reagent test, it is not an aldehyde but it must be a ketone.

Since it doesn’t give iodoform test, it doesn’t have group.
Hence the structure of ‘A’ will be,

Question 142.
Identify A and B in the following reactions :

Answer:

 

Answer:

 

Answer:

 

Answer:

Question 143.
What is the action of concentrated nitric acid on (1) Benzaldehyde (2) Benzophenone?
Answer:

1. Benzaldehyde on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzaldehyde
   
2. Benzophenone on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzophenone
   

Question 144.
Explain laboratory tests for carboxyl (- COOH) group.
Answer:
The presence of – COOH group in carboxylic acids is identified by the following tests :

(1) Litmus test : (valid for water soluble substances)
Aqueous solution of Organic compound containing – COOH group turns blue litmus red which indicates the presence of acidic functional group.

(2) Sodium bicarbonate test : When sodium bicarbonate is added to an organic compound containing – COOH group, a brisk effervescence of carbon dioxide gas is evolved. Water insoluble acid goes in solution and gives precipitate an acidification with cone. HCl. This indicates the presence of -COOH group.

(3) Ester test : One drop of concentrated sulfuric acid is added to a mixture of given organic compound containing – COOH group and one mL of ethanol, the reaction mixture is heated for 5 minutes in hot water bath. After this, hot solution is poured in a beaker containing water, fruity smell of ester confirms the presence of carboxylic acid.

Question 145.
How is acid chloride obtained from carboxylic acid?
Answer:
Carboxylic acid on heating with thionyl chloride (SOCl₂), phosphorus trichloride (PCl₃) or phosphorus pentachloride (PCl₅) give corresponding acid chlorides. In this reaction – OH of carboxyl group is replaced by -Cl.

The reactions are :
(1) Action on SOCl₂ on carboxylic acid :

Example : Acetic acid reacts with thionyl chloride to give acetyl chloride.

(2) Action of PCl₃ on carboxylic acid (ethanoic acid) :

Example : Action of phosphorus trichloride on acetic acid gives acetyl chloride.

(3) Action of PCI₅ on carboxIic acid :

Question 146.
How will you convert 3,5-Dinitrobenzoic acid to 3,5-Dinitrobenzoyl chloride?
Answer:
When 3,5-Dinitrobenzoic acid is heated with phosphorus pentachloride, 3,5-Dinitrobenzoyl chloride is obtained.

Question 147.
How is acid amide obtained from carboxylic acid?
Answer:
Carboxylic acid or acid chloride with ammonia salts, which on further strong heating at high temperature decompose to give amides.

When acetic acid is treated with ammonia, ammonium acetate is obtained. Ammonium acetate on strong heating decomposes to form acetamide.

When acetyl chloride is treated with ammonia, acetamide is obtained

Question 148.
How is acid anhydride obtained from carboxylic acid?
Answer:
When carboxylic acid is heated with strong dehydrating agent like phosphorus pentoxide or concentrated sulphuric acid, an acid anhydride is obtained,

The reaction is reversible and anhydride is hydrolysed back to acid.

Alternatively, when sodium acetate is heated with acetyl chloride, acetic anhydride is obtained. This reaction is irreversible.

Question 149.
What is decarboxylation of an acid ? How is it done?
OR
What happens when sodium acetate is heated with soda lime ?
Answer:
Removal of a carboxylic group from acid is called decarboxylation. Decarboxylation of an acid is carried out by heating anhydrous sodium salts of carboxylic acids with soda lime (NaOH + CaO). The product hydrocarbons obtained contain one carbon atom less than the carboxylic acid.

When sodium acetate is heated with soda lime, methane is obtained.

Question 150.
How is alcohol obtained from carboxylic acid ?
OR
What is the action of LiAlH₄/H₃O⁺on ethanoic acid? Write balanced equation for the conversion :
Cyclopropane carboxylic acid to Cyclopropylmethanol.
Answer:
Carboxylic acids are reduced to primary alcohols using powerful reducing agent lithium aluminium hydride.

(i) When ethanoic acid is reduced in the presence of LiAlH4 in dry ether, forms ethanol.

(ii) When cyclopropane carboxylic acid is reduced in the presence of lithium aluminium hydride in dry ether, forms cyclopropyl methanol.

Question 151.
What is the action 9f following compounds on cyclohexanone in presence of dry hydrogen chloride?
(1) Ethyl alcohol
(2) Ethylene glycol
Answer:
(1) With Ethyl alcohol : Cyclohcxanonc reacts with one equivalent of monohydric ethyl alcohol lo form hemi ketal, which further adds another molecule of alcohol to form a gem-dialkox compound known as ketal.

(2) With Ethylene glycol : cyclohexanone reaCts with ethylene glycol to form cyclic ketal.

Question 152.
Answer the following in one sentence.

1. Name the compound which reacts with formaldehyde to produce ethyl alcohol.
Answer:
The compound which reacts with formaldehyde to produce ethyl alcohol is methyl magnesium iodide.

2. What are imines ?
Answer:
These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with primary amine.

3. Why does skin have burning sensation, when an ant bites ?
Answer:
When an ant bites, formic acid is released from an ant which gives burning sensation as the acid comes in contact with the skin.

4. What is the percentage of acetic acid in vinegar?
Answer:
The percentage of acetic acid in vinegar is 6 to 8%.

5. Which reagent is used to distinguish formic acid and acetic acid?
Answer:
The reagent used to distinguish formic acid and acetic acid is ammoniacal silver nitrate.

6. What happens when acetyl chloride is treated with dibenzyl cadmium ? Give reaction.
Answer:

7. Complete the following reaction :

Answer:

8. Give reason : Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
Answer:
Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the catalyst aluminium chloride (Lewis acid) gets bonded to carboxyl group.

9. Write the name of two compounds which do not contain carbonyl group but show iodoform test.
Answer:
The name of two compounds which do not contain carbonyl group but show iodoform test are ethanol

10. Give reason : In semicarbazide, – NH₂ group bonded to carbonyl group is not involved in the formation of semicarbazone.
Answer:
NH₂ group attached to – NH group in semicarbonide is more active than NH2 group attached to carbonyl group due to electron density difference.

11. Fehling solution does not oxidise benzaldehyde but Tollen’s reagent oxidises benzaldehyde. Give reason.
Answer:
When benzaldehyde is heated with Fehling solution, there is no change in colour of the solution, Cu²⁺ ion is not reduced, hence Fehling solution does not oxidise benzaldehyde. However, Tollen’s reagent oxidises benzaldehyde to give silver mirror test.

12. Give reason : Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids.
Answer:
Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids due to greater electronegativity of sp²-hybridised carbon to which carboxyl carbon is attached.

Multiple Choice Questions

Question 153.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. IUPAC name of

(a) l-Phenylhexan-2-one
(b) 6-Phenylhexan-5-one
(c) l-Benzylhexan-2-one
(d) Dodecan-5-one
Answer:
(a) l-Phenylhexan-2-one

2. The general formula of carbonyl compounds is
(a) C_nH_2n+1OH
(b) C_nH_2nO
(c) C_nH_2nO₂
(d) C_nH_2n+1O
Answer:
(b) C_nH_2nO

3. Aldehydes and ketones are
(a) chain isomers
(b) functional isomers
(c) geometrical isomers
(d) position isomers
Answer:
(b) functional isomers

4. Identify ‘C’ in the following reaction,
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{H}^{+} \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{C}\)
(a) Propan – 1 – ol
(b) Propanone
(c) 2-Ethyl-3-pentanone
(d) Propanal
Answer:
(d) Propanal

5. Grignard reagent when reacted with alkyl cyanide followed by hydrolysis gives
(a) an aldehyde
(b) a ketone
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(b) a ketone

6. Identify ‘ B ’ in the following reaction :
\(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{N} \stackrel{\mathrm{CH}_{3} \mathrm{MgI}}{\longrightarrow} \mathrm{A} \underset{\mathrm{HCl}}{\stackrel{2 \mathrm{HOH}}{\longrightarrow}} \mathrm{B}+\mathrm{NH}_{3}+\mathrm{MgIOH}\)
(a) Magnesium intermediate
(b) Ethanol
(c) Propanal
(d) Propanone
Answer:
(d) Propanone

7. \(\mathrm{A}+\mathrm{B} \stackrel{\text { dry ether }}{\longrightarrow} \mathrm{Col}\) Complex \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}\) Ethyl Methyl Ketone. In the above reaction, A and B are
(a) Formonitrile, Propyl magnesium bromide
(b) Ethyl cyanide, Ethyl magnesium bromide
(c) Hydrogen cyanide, Ethyl magnesium bromide
(d) Acetonitrile, Ethyl magnesium bromide
Answer:
(d) Acetonitrile, Ethyl magnesium bromide

8. A dilute solution of p-rosaniline hydrochloride in water whose pink colour has been discharged by passing sulphur dioxide, does not restore its colour by
(a) HCHO
(b) CH₂CHO
(c) (CH₃)₂COCH₃
(d) CCl₃CHO
Answer:
(c) (CH₃)₂COCH₃

9. The reagent with which both acetaldehyde and acetone reacts easily is –
(a) Fehling’s solution
(b) Tollen’s reagent
(c) Grignard reagent
(d) Schiff s reagent
Answer:
(c) Grignard reagent

10. Isopropyl methyl ketone when treated with Zn-Hg and concentrated hydrochloric acid give
(a) iso-butane
(b) iso-pentane
(c) n-pentane
(d) neo-pentane
Answer:
(b) iso-pentane

11. The formation of acetone cyanohydrin from acetone is an example of
(a) Nucleophilic addition
(b) Nucleophilic substitution
(c) Electrophilic addition
(d) Electrophilic substitution
Answer:
(a) Nucleophilic addition

12. Which of the following is Fehling solution ‘A’?
(a) CuSO₄ solution
(b) CaSO₄ solution
(c) NaOH solution
(d) Sodium potassium tartarate solution
Answer:
(a) CuSO₄ solution

13. The compound ‘X’ upon alkaline hydrolysis gives a product which reacts with phenylhydrazine but does not reduce ammoniacal silver nitrate solution. A possible structure for ‘X’ is
(a) CH₃CHCl CH₂Cl
(b) CH₃CCl₂CH₃
(c) CH₃CH₂CH₂Cl
(d) CH₃CH₂CHCl₂
Answer:
(b) CH₃CCl₂CH₃

14. Which of the following is the correct statement with respect to aldehyde and ketones ?
(a) Ketones are reducing agents
(b) Aldehydes are good reducing agents
(c) Cannizzaro reaction is an addition reaction
(d) Ketones do not react with Grignard reagent
Answer:
(b) Aldehydes are good reducing agents

15. Acetaldehyde acts as
(a) a catalyst
(b) a reducing agent
(c) an oxidizing agent
(d) a mordant
Answer:
(b) a reducing agent

16. An organic compound (A) C₃H₈0 on oxidation gives (B) C₃H₆O₂. The compound A may be
(a) an ester
(b) an alcohol
(c) an aldehyde
(d) a ketone
Answer:
(b) an alcohol

17. Identify ‘B’ from the following reaction :
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NH}_{2} \mathrm{OH} \rightarrow \mathrm{A} \stackrel{\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\Delta} \mathrm{B}\)
(a) Propan-1-amine
(b) Propan-2-amine
(c) Isopropylamine
(d) Dimethylamine
Answer:
(a) Propan-1-amine

18. Sodium borohydride does not reduce
(a) – COOH group
(b) – NO₂ group
(c) – X atom
(d) – CHO group
Answer:
(a) – COOH group

19. An aldehyde when warmed with Zn/Hg and cone. HCl gives
(a) alcohol
(b) hydrocarbon
(c) carboxylic acid
(d) ketone
Answer:
(b) hydrocarbon

20. Acetaldol is
(a) 3-hydroxy butanol
(b) 3-hydroxy butanal
(c) 2-hydroxy propanal
(d) 3-hydroxy pentanal
Answer:
(b) 3-hydroxy butanal

21. Acetone can be reduced to propane, the reduction is called
(a) Clemmensen’s reduction
(b) catalytic reduction
(c) Rosenmund’s reduction
(d) partial reduction
Answer:
(a) Clemmensen’s reduction

22. Which of the following reagents can react with acetaldehyde to give water soluble white crystal-line solid?
(a) NaHSO₄
(b) NaHSO₃
(c) Na₂SO₃
(d) Na₂SO₄
Answer:
(b) NaHSO₃

23. Which of the following compounds does NOT undergo aldol condensation?

Answer:
(a)

24. Formalin is 40% aqueous solution of :
(a) Methanal
(b) Methanoic acid
(c) Methanol
(d) Methanamine
Answer:
(a) Methanal

25. Which of the following compounds do not produce pink colour with Schiff s reagent?
(a) Formaldehyde
(b) 2-propanone
(c) 3-pentanone
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

26. Both aldehydes and ketones can react with
(a) Tollen’s reagent
(b) the Grignard reagent
(c) Fehling’s solution
(d) Schiffs reagent
Answer:
(b) the Grignard reagent

27. Aldol reaction is.
(a) an addition reaction
(b) an elimination reaction
(c) a self-reduction reaction
(d) a disproportionate ion reaction
Answer:
(a) an addition reaction

28. The reaction in which two molecules combine to form a new molecule with the elimination of a small molecule like water is called
(a) an oxidation reaction
(b) a condensation reaction
(c) a hydrolysis reaction
(d) a redox reaction
Answer:
(b) a condensation reaction

29. Benzaldehyde undergoes Cannizzaro’s reaction to give
(a) sodium benzoate and methyl alcohol
(b) sodium benzoate and benzyl alcohol
(c) benzyllic acid and benzyl alcohol
(d) phenol and benzoic acid
Answer:
(b) sodium benzoate and benzyl alcohol

30. Identify ‘X’ in the following reaction :
CH₃-CHO + X → CH₃-CH = N-NH-C₆H₅ + H₂O
(a) C₆H₅-NH₂
(b) C₆H₅-NH-NH₂
(c) C₆H₅-N = NH
(d) C₆H₅-NH-NH-CH₃
Answer:
(b) C₆H₅-NH-NH₂

31. What happens when propanal is treated with zinc amalgam and conc.HCl ?
(a) Propan-l-ol
(b) Propan-2-ol
(c) Propane
(d) Propanone
Answer:
(c) Propane

32. Identify ‘ B ’ in the following reaction :
\(2 \mathrm{CH}_{3}-\mathrm{CHO} \frac{\text { dil. base or acid }}{300 \mathrm{~K}} \mathrm{~A} \underset{\text { dehydration }}{\text { dehy }} \mathrm{B}+\mathrm{H}_{2} \mathrm{O}\)
(a) CH₃-CH(OH)-CH₂-CHO
(b) CH₃-CH₂-CH(OH)-CHO
(c) CH₃-CH = CH-CHO
(d) CH₃-CO-CH₃
Answer:
(c) CH₃-CH = CH-CHO

33. The blue colour of Fehling’s solution is due to
(a) Cu₂O
(b) CuCO₃
(c) CuO
(d) Cu⁺⁺ ions
Answer:
(d) Cu⁺⁺ ions

34. How is Schiff s reagent prepared?
(a) By passing CO₂ through p-rosaniline solution
(b) By passing NO₂ through p-rosaniline solution
(c) By passing SO₂ through p-rosaniline solution
(d) By passing NH₃ through silver nitrate solution
Answer:
(c) By passing SO₂ through p-rosaniline solution

35. Benzaldehyde when treated with cone. HNO₃ gives
(a) o-nitrobenzaldehyde
(b) p-nitrobenzaldehyde
(c) m-nitrobenzaldehyde
(d) a mixture of -o and -p-nitrobenzaldehyde
Answer:
(c) m-nitrobenzaldehyde

36. Which of the following carbonyl compounds undergoes aldol condensation ?
(a) Benzaldehyde
(b) Benzophenone
(c) Acetophenone
(d) tert-Butyl phenyl ketone
Answer:
(c) Acetophenone

37. Which of the following carbonyl compounds undergoes self redox reaction in presence of concentrated base ?
(a) 3-Methylpentanal
(b) 2-Chlorobutanal
(c) 2,2-Dimethylpropanal
(d) tert-Butyl methyl ketone
Answer:
(c) 2,2-Dimethylpropanal

38. The smell of bitter almond is given by the compound.
(a) Benzoic acid
(b) Benzaldehyde
(c) Vanillin
(d) Cinnamaldehyde
Answer:
(b) Benzaldehyde

39. Which of the following will not give yellow precipitate when treated with NaOH and H?
(a) 3-Methylbutan-2-one
(b) 2-methylpentan-3-one
(c) Propanone
(d) Hexan-2-one
Answer:
(b) 2-methylpentan-3-one

40. A β-hydroxyl carbonyl compound is obtained by the action of NaOH on
(a) HCHO
(b) C6H₅CHO
(c) CR₃CHO
(d) CH₃CHO
Answer:
(d) CH₃CHO

41. Decarboxylation of sodium propionate gives
(a) methane
(b) ethane
(c) propane
(d) ethene
Answer:
(b) ethane

42. Ester on hydrolysis with dil HCl gives
(a) RCOOH + ROH
(b) RCOR + ROH
(c) ROH + ROH
(d) RCOR + RCOOH
Answer:
(a) RCOOH + ROH

43. \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3} \stackrel{\mathrm{CrO}_{3}}{\longrightarrow} \mathrm{A}\) The compound A is
(a) acetic acid
(b) propionic acid
(c) formic acid
(d) benzoic acid
Answer:
(a) acetic acid

44. The reaction of C₆H₅CH = CHCHO with LiAlH₄ gives
(a) C₆H₅CH₂CH₂CH₂OH
(b) C₆H₅CH₂CH₂CHO
(c) C₆H₅CH = CHCH₂OH
(d) C₆H₅CH₂CHOHCH₃
Answer:
(a) C₆H₅CH₂CH₂CH₂OH

45. A mixture of sodium benzoate and sodalime on heating yields
(a) methane
(b) benzene
(c) sodium benzoate
(d) calcium benzoate
Answer:
(b) benzene

46. Which is the strongest acid?
(a) CH₃COOH
(b) CH₃CH₂COOH
(c) (CH₃)₃CCOOH
(d) CICH₂COOH
Answer:
(d) CICH₂COOH

47. Benzaldehyde when treated with alkaline KMnO₄ yields
(a) Benzyl alcohol
(b) Benzoic acid
(c) CO₂ and H₂O
(d) Salicylic acid
Answer:
(a) Benzyl alcohol

48. Acetonitrile on acidic hydrolysis gives
(a) HCOOH
(b) CH₃NC
(c) CH₃COONa
(d) CH₃COOH
Answer:
(d) CH₃COOH

49. The organic compounds A and B reacts with sodium metal and liberates hydrogen gas. A and B reacts together to give ethyl acetate. The A and B are
(a) CH₃COOH & C₂H₅OH
(b) HCOOH & C₂H₅OH
(c) CH₃COOH & HCOOH
(d) CH₃COOH & CH₃OH
Answer:
(a) CH₃COOH & C₂H₅OH

50. Which one of the following undergoes reaction with 50% NaOH to give to corresponding alcohol and acid?
(a) Phenol
(b) Benzoic acid
(c) Benzaldehyde
(d) Butanal
Answer:
(c) Benzaldehyde

51. Identify the reactant in the following reaction.

Answer:
(c) CO

52. The strongest acid is

Answer:
(c)

53. Predict the product in the following reaction

The compound A is
(a) butane
(b) propane
(c) ethane
(d) propene
Answer:
(b) propane

54. Ethyl benzoate when heated with dil H₂SO₄ gives
(a) acetic acid
(b) benzoic acid
(c) ethanoic acid
(d) phenyl methanol
Answer:
(b) benzoic acid

55. Margarine contains
(a) acetaldehyde
(b) propionaldehyde
(c) butyraldehyde
(d) formaldehyde
Answer:
(c) butyraldehyde

56. Monocarboxylic acids have the general formula
(a) C_nH_2n+1O₂
(b) C_nH_2nO₂
(c) C_nH_2nO
(d) C_nH_2n-1O₂
Answer:
(b) C_nH_2nO₂

57. Formic acid is obtained from
(a) vinegar
(b) red ants
(c) butter
(d) oil
Answer:
(b) red ants

58. Butter contains
(a) lactic acid
(b) butyric acid
(c) citric acid
(d) acetic acid
Answer:
(b) butyric acid

59. Glacial acetic acid is
(a) HCOOH
(b) CH₃COOH
(c) CH₃CH₂COOH
(d) C₃H₇COOH
Answer:
(b) CH₃COOH

60. Which of the following acids is optically active?
(a) Oxalic acid
(b) Salicylic acid
(c) Acetic acid
(d) Lactic acid
Answer:
(d) Lactic acid

61. Lactic acid is
(a) propionic acid
(b) α-hydroxy propionic acid
(c) p-hydroxy benzoic acid
(d) butyric acid
Answer:
(b) α-hydroxy propionic acid

62. The carbon atom of the carboxylic group is
(a) sp³-hybridized
(b) sp²-hybridized
(c) sp-hybridized
(d) unhybridized
Answer:
(b) sp²-hybridized

63. The common name of carboxylic fatty acids is derived from
(a) the name of parent alkanes
(b) the name of corresponding aldehydes
(c) from their original sources
(d) the name of alkyl group present in them
Answer:
(c) from their original sources

64. The IUPAC name of a-methylpropionic acid is
(a) Propanoic acid
(b) Butanoic acid
(c) 2-Methylpropanoic acid
(d) 2-Methylbutanoic acid
Answer:
(c) 2-Methylpropanoic acid

65. For the nomenclature of carboxylic acids, the suffix used is
(a) -ane
(b) -oic
(c) -al
(d) -ol
Answer:
(b) -oic

66. Propionic acid can be prepared by the
(a) action of propyl magnesium chloride on dry ice
(b) alkaline hydrolysis of propyl cyanide
(c) acid hydrolysis of ethyl cyanide
(d) oxidation of Propanone
Answer:
(c) acid hydrolysis of ethyl cyanide

67. The intermediate compound formed during hy-drolysis of acetonitrile to acetic acid is
(a) acetone
(b) acetamide
(c) ammonium acetate
(d) ethyl ammonium chloride
Answer:
(b) acetamide

68. Carbonation of CH₃MgI gives organic compound. The same compound can also be obtained by
(a) oxidation of Methanol
(b) oxidation of Methanal
(c) acid hydrolysis of acetonitrile
(d) alkaline hydrolysis of ethyl cyanide
Answer:
(c) acid hydrolysis of acetonitrile

69. The acid that cannot be prepared by the action of Grignard reagent on dry ice is
(a) methanoic acid
(b) ethanoic acid
(c) propanoic acid
(d) butanoic acid
Answer:
(a) methanoic acid

70. The compound which on acid hydrolysis followed by oxidation gives acetic acid is
(a) CH₃I
(b) CH₂Cl₂
(c) ClCH₂CH₂C1
(d) CH₃CHCl₂
Answer:
(d) CH₃CHCl₂

71. The hydrolysis product of alkyl cyanide is
(a) primary amine
(b) amides
(c) aldehyde
(d) carboxylic acid
Answer:
(d) carboxylic acid

72. To prepare acetic acid,
(a) methyl alcohol is oxidized with KMnO₄
(b) calcium acetate is distilled with calcium for-mate under dry conditions
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄
(d) glycerol is heated with H₂SO₄
Answer:
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄

73. Solid carbon dioxide when treated with etheral solution of C₂H₅MgBr followed by acid hydrolyzis gives
(a) propanoic acid
(b) ethanoic acid
(c) propionic acid
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

74. Which of the following compound does not give acetic acid on oxidation ?
(a) Ethanol
(b) Propan-l-ol
(c) Propan-2-ol
(d) 2-Methyl propan-2-ol
Answer:
(b) Propan-l-ol

75. A carboxylic acid resembles an alcohol with respect to its reaction with
(a) acidified K₂Cr₂O₇
(b) washing soda
(c) caustic soda
(d) sodium metal
Answer:
(d) sodium metal

76. Acetic acid can be converted into acetic anhydride on heating with
(a) POCl₃
(b) PCl₃
(c) PCI₅
(d) P₂O₅
Answer:
(d) P₂O₅

77. Acetyl chloride reacts with ammonia to give
(a) ammonium acetate
(b) ethylammonium chloride
(c) ethylamine
(d) acetamide
Answer:
(d) acetamide

78. The reagent that reacts with acetic acid to give sodium acetate with liberation of carbon dioxide gas is
(a) sodium metal
(b) caustic soda
(c) caustic potash
(d) baking soda
Answer:
(d) baking soda

79. An alkene on hydration gives a compound, which reacts with propionic acid to produce isopropyl propionate. The alkene is
(a) CH₂ = CH₂
(b) CH₃-CH = CH₂
(c) CH₃ – CH₂ – CH = CH₂
(d) CH₃ – CH = CH – CH₃
Answer:
(b) CH₃-CH = CH₂

80. Both the compounds ‘A’ and ‘B’ react with sodium metal to liberate hydrogen gas and react with each other to give Methylethanoate. The compounds ‘A’ and ‘B’ are
(a) C₂H₅ – COOH and CH₃ – OH
(b) C₂H₅ – COOH and C₂H₅ – OH
(c) CH₃ – COOH and C₂H₅ – OH
(d) CH₃ – COOH and CH₃ – OH
Answer:
(d) CH₃ – COOH and CH₃ – OH

81. Identify the product ‘D’ in the following series of reactions.

(a) CH₃COOCH₃
(b) CH₃COOC₂H₅
(c) C₂H₅COOCH₃
(d) C₂H₅COOC₂H₅
Answer:
(b) CH₃COOC₂H₅

82. Acetyl chloride on heating with sodium acetate gives
(a) ethyl acetate
(b) acetamide
(c) acetic anhydride
(d) acetaldehyde
Answer:
(c) acetic anhydride

83. Carboxylic acids are acidic in nature because
(a) it dissociates to give H⁺ ions
(b) it donates proton
(c) it reacts with active metal and liberates hydro-gen gas
(d) all of these
Answer:
(d) all of these

84. Carboxylic acids on heating with P₂O₅ give
(a) acid chlorides
(b) alkyl halides
(c) acid amides
(d) acid anhydrides
Answer:
(d) acid anhydrides

85.

Answer:
(c)

86. The compound having general formula is called

(a) diester
(b) acid anhydride
(c) hemiacetal
(d) acetal
Answer:
(d) acetal

87. Identify the strongest acid amongst the following :
(a) Chloroacetic acid
(b) Acetic acid
(c) Trichloroacetic acid
(d) Dichloroacetic acid
Answer:
(c) Trichloroacetic acid

88. Acetaldehyde, when treated with which among the following reagents does ‘not’ undergo addition reaction?
(a) ammonia
(b) hydroxyl amine
(c) ammoniacal silver nitrate
(d) semicarbazide
Answer:
(c) ammoniacal silver nitrate

89. Popcorn has butter flavour which contains
(a) butan-l-one
(b) butane-2, 3-dione
(c) butan-2-one
(d) butyric acid
Answer:
(b) butane-2, 3-dione

90. On acid hydrolysis, propane nitrile gives
(a) propanal
(b) acetic acid
(c) propionamide
(d) propanoic acid
Answer:
(d) propanoic acid

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 9 Business Communication Skills of a Secretary Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 9 Business Communication Skills of a Secretary

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Lay out of a letter means ____________
(a) physical appearance
(b) arrangement of its parts
(c) body
Answer:
(b) arrangement of its parts

Question 2.
Written communication is called ____________
(a) compliments
(b) consideration
(c) correspondence
Answer:
(c) correspondence

Question 3.
Communication that takes place internally between various departments of an organization ____________
(a) external communication
(b) internal communication
(c) verbal communication
Answer:
(b) internal communication

Question 4.
Communication that takes place between business organization and outsiders like bank, suppliers, creditors, etc ____________
(a) internal communication
(b) external communication
(c) verbal communication
Answer:
(b) external communication

Question 5.
Communication conveying message in spoken form is called ____________
(a) Non-verbal communication
(b) Verbal communication
(c) Internal communication
Answer:
(b) Verbal communication

Question 6.
Facial expression is a form of ____________ communication.
(a) Verbal communication
(b) Non-verbal communication
(c) Internal communication
Answer:
(b) Non-verbal communication

Question 7.
Information or ideas in written form is ____________
(a) oral communication
(b) written communication
(c) body language
Answer:
(b) written communication

Question 8.
A website for publishing informal online articles is called ____________
(a) blog
(b) cellular phone
(c) video conferencing
Answer:
(a) blog

Question 9.
Two-way personalised webscam based communication is called as ____________
(a) website
(b) blog
(c) video conferencing
Answer:
(c) video conferencing

Question 10.
____________ is a skill of being able to understand the feelings of another person.
(a) Empathy
(b) Sympathy
(c) Body language
Answer:
(a) Empathy

Question 11.
A ____________ introduces the firm.
(a) inside
(b) address, heading
(c) reference number
Answer:
(b) address, heading

Question 12.
Language of business letter should be ____________
(a) simple
(b) hard
(c) flowery
Answer:
(a) simple

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Correspondence	(1) Greeting to the recipient of the letter
(b) Salutation	(2) Message not included in the body of the letter
(c) Postscript	(3) Non-verbal communication
(d) Hand gestures	(4) Introduction of the sender
(e) Conciseness	(5) Written communication
	(6) Inside address
	(7) Brief matter
	(8) Other documents attached
	(9) Verbal communication
	(10) Heading

Answer:

Group ‘A’	Group ‘B’
(a) Correspondence	(5) Written communication
(b) Salutation	(1) Greeting to the recipient of the letter
(c) Postscript	(2) Message not included in the body of the letter
(d) Hand gestures	(3) Non-verbal communication
(e) Conciseness	(7) Brief matter

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Branch of general communication concerned with business activities.
Answer:
Business communication

Question 2.
Communication that takes place internally between departments.
Answer:
Internal communication

Question 3.
Communication that takes place between organization and outsider.
Answer:
External communication

Question 4.
An electronic mail sending messages using electronic devices through the internet.
Answer:
E-mail

Question 5.
A website for publishing informal articles.
Answer:
Blog

Question 6.
Online interactive groups using advanced mobile and web-based technologies.
Answer:
Social media network

Question 7.
The gesture of a dancer.
Answer:
Non-verbal communication

Question 8.
Skill to understand and share the feelings of another person.
Answer:
Empathy

Question 9.
The name and address of the letter writer.
Answer:
Heading

Question 10.
A matter wrote after completing the letter.
Answer:
Postscript

Question 11.
Copies were sent along with the letter.
Answer:
Enclosure

Question 12.
The final part of the letter.
Answer:
Signature

1D. State whether the following statements are True or False.

Question 1.
A report is a systematic presentation of facts, figures, and conclusions about a topic.
Answer:
True

Question 2.
A notice is a written summary of transactions conducted at the meeting.
Answer:
False

Question 3.
Minutes give precise information regarding an important event that is about to take place.
Answer:
False

Question 4.
Letterhead includes the name and address of the sender.
Answer:
True

Question 5.
A letter written in a logical sequence is coherence.
Answer:
True

Question 6.
“You’ attitude means writing the word you many times in a letter.
Answer:
False

Question 7.
Harsh and rude words should be avoided in a letter.
Answer:
True

Question 8.
Minimum words should be used in a letter.
Answer:
True

Question 9.
The additional information written after the letter is completed is Postscript.
Answer:
True

Question 10.
The letter sent to other people at the same time is ‘Carbon Copy Notation’.
Answer:
True

Question 11.
Salutation includes documents, cheques, etc. attached with the letter.
Answer:
False

Question 12.
A complimentary close is written below the body of the letter in a polite manner.
Answer:
True

1E. Find the odd one.

Question 1.
Twitter, Youtube, Facebook, Written communication
Answer:
Written communication

Question 2.
E-mail, website, blog, completeness
Answer:
Completeness

Question 3.
Notice, Reports, Minutes, Non-verbal communication
Answer:
Non-verbal communication

Question 4.
Coherence, Consideration, Cheerfulness, Margin
Answer:
Margin

1F. Complete the sentences.

Question 1.
The process of communication conveying a message in spoken form is called as ____________
Answer:
verbal communication

Question 2.
Communication which is neither written or spoken is known as ____________
Answer:
Non-verbal communication

Question 3.
The participants are able to see and hear each other and also display visual data models etc. under ____________
Answer:
video conferencing

Question 4.
Communication that involves hearing and understanding what a person says to you is ____________
Answer:
active listening

Question 5.
Proper arrangement of various parts of the letter is called as ____________
Answer:
layout

Question 6.
The number written on the left-hand side below the heading to give a quick reference to the matter concerned is called ____________
Answer:
Reference Number

Question 7.
The reader gets the idea of the matter of the letter without reading the letter completely by reading ____________
Answer:
Subject

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Written summary	……………………….
(2) Spirit of hopeness	………………………..
(3) Logical sequence	……………………….
(4) ……………………….	Your’s faithfully

(Coherence, Minutes, Cheerfulness, Complimentary close)
Answer:

Group ‘A’	Group ‘B’
(1) Written summary	Minutes
(2) Spirit of hopeness	Cheerfulness
(3) Logical sequence	Coherence
(4) Your’s faithfully	Your’s faithfully

1H. Answer in one sentence.

Question 1.
Name the Electronic device in which mails can be sent.
Answer:
Mails can be through E-mail (Internet).

Question 2.
Name the device that provides a short message service.
Answer:
Mobile/Cellular phone provides short message service.

Question 3.
What is the final part of the letter called?
Answer:
The final part of the letter is called ‘Signature’.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Written communication includes body language, facial expression, etc.
Answer:
Non Verbal communication includes body language, facial expression, etc.

Question 2.
Internal communication takes place between business organizations and outsiders.
Answer:
External communication takes place between business organizations and outsiders.

Question 3.
Coherence is also called the use of ‘You attitude’.
Answer:
Consideration is also called the use of ‘You attitude’.

Question 4.
A written summary of a business transacted at the meeting is called Notice.
Answer:
A written summary of a business transacted at the meeting is called Minutes.

Question 5.
Oral Communication is a permanent record.
Answer:
Written Communication is a permanent record.

Question 6.
The skill of understanding the feelings of another person is called Sympathy.
Answer:
The skill of understanding the feelings of another person is called Empathy.

Question 7.
A letter without clarity is invalid.
Answer:
A letter without a signature is invalid.

2. Explain the following terms/concepts.

Question 1.
Non-Verbal Communication
Answer:

• Communication that does not involve written or spoken words is called non-verbal communication.
• It takes place by using body language, facial expression, eye contact, silence, symbols, signs, gestures, etc.

Question 2.
Notice
Answer:

• It is an intimation by the company to the concerned persons or members or directors about the day, date, time, place, and business to a transacted at the meeting.
• It is to be sent to all the concerned persons through registered post.
• It should also include the agenda which is going to be discussed at the meeting.

3. Answer in brief.

Question 1.
Explain any four modes of Electronic devices.
Answer:
Modes of electronic device:
(a) E-mail:
It means sending messages through the internet. An E-mail has made businesses come closer to their customers. It also saves time for writing letters, posting a letter, etc.

(b) Websites:
It is a set of interconnected web pages which are located on a single web. It contains the information provided by the owner of the website. It can be accessed through the internet or through a private local area network. Each website has a unique internet address called “Uniform Resource Locator”(URL).

(c) Social media network:
It is a very popular online interactive group created by using mobile and web-based technologies. It helps the business organization to interact with the consumers and communicate about their products and services.

(d) Video-conferencing:
It can be done through computers by providing a video link between two or more people. The participants can see and hear each other while communicating and can also display visual data also.

4. Justify the following statements.

Question 1.
The letterhead introduces the firm.
Answer:

• The letterhead (or the head address) consists of the name of the company sending out the letter, its registered address, its emblem or logo, telephone number, fax number, telegraphic address, telex number, website, and e-mail address, and same times a brief line describing the nature of business of the company.
• The letterhead is usually printed at the top center of the page.
• When the addressee of the letter receives the letter, his attention first goes to the letterhead.
• On seeing the letterhead, he knows about the sender’s firm.
• The layout of the letterhead, it’s printing, and its style help to create an image of the company.
• An attractive letterhead creates a good impression.
• The letterhead speaks a lot of the sender’s organization.
• Thus, the letterhead introduces the firm.

Question 2.
Postscript should not be used in business letters.
Answer:

• Postscript means ‘written afterward’. It is like a footnote added to the letter.
• It means additional matter not written in the main body of the letter but added after completing the letter.
• It is necessary if some information is received late and is required to be conveyed along with the original letter.
• To have a postscript in the letter is not a good practice and hence should be avoided.
• It indicates the possible carelessness of the writer.
• It should be used only for communicating an urgent message.
• Thus, postscript should not be used in business letters.

Question 3.
Courtesy is one of the requisites of a good business letter.
Answer:

• Every business letter must be courteously worded.
• The language used should be polite, pleasing, and convincing.
• Courtesy and polite tone are possible by using terms like ‘please’, ‘thank you, ‘very kind of you’ etc.
• It is specifically necessary for letters sent to the employees.
• It includes polite manners, kindness, and consideration for others.
• A courteous letter builds goodwill for the organization.
• Lack of courtesy may prove to be costly and troublesome to a business unit.
• A business letter and serves as a ‘silent salesman’ or ‘silent ambassador’ only when it is courteous in meaning and language.
• Courtesy facilitates the expansion of business activities.
• Thus, courtesy is one of the requisites of a good business letter.

Question 4.
A business letter is a silent salesman.
Answer:

• In modern times, the scale of the business has increased to such an extent that direct and personal dealing with the customers is difficult except in case of retail trade.
• Correspondence is, thus, necessary to widen the boundaries of markets and expand the scale of business.
• Effectively drafted and neatly typed letters play the role of a salesman in creating and expanding the market for goods and services.
• When letters are courteous it creates a good impression.
• Letters convince the prospects of the novelty, specialty, and utility of the products and induce them to place orders.
• Thus, a business letter is a silent salesman.

Question 5.
A letter without a date is incomplete.
Answer:

• The date is an important part of a business letter.
• The date includes the day, month, and year on which the letter is sent to the addressee.
• The reference of data is useful for the fulfillment of business transactions on a particular date.
• A letter with a date can serve as evidence in a court of law.
• The date on the letter facilitates quick reference to old letters and filing of letters in chronological order.
• A letter without a date is incomplete and legally invalid. It is like a body without ahead.
• Thus, a letter without a date is incomplete.

Question 6.
The margin on all sides is a waste of paper.
Answer:

• In order to make a letter appear systematic and attractive, it is better to maintain margins on both sides of the letter.
• Proper margin makes the letter attractive.
• Proper margin must be kept on the left-hand side and right-hand side.
• Proper space must be left at the bottom of the letter.
• Thus, the margin on all sides is a waste of paper.

Question 7.
The signature provides legal value to the letter.
Answer:

• It usually consists of the name of the writer, his official designation, if any, and the department he is concerned with.
• It is placed directly below the complimentary close.
• The signature must be handwritten in ink and placed two or three line spaces below the complimentary close.
• Business letters, today, carry the typed name and also the designation of the signatory below the signature.
• This helps the reader to identify the person by his name and his official position in the company.
• A letter without a date is invalid, likewise, a letter without a signature is also invalid.
• The unsigned letter has no significance.
• Thus, a signature provides legal value to the letter.

5. Answer the following questions.

Question 1.
Explain the physical appearance of a letter.
Answer:
‘Physical appearance’ simply means ‘The look’ of the letter. How it should appear or present itself to the receiver of the letter, what impression should it create? All these depend on the ‘Look’ or ‘Outer appearance’ of the letter. Therefore to make the letter look attractive, attention should be given to the following points.

(a) Paper:
A paper of superior quality should be used. The high cost of such paper is a sort of investment and not an expenditure. A paper of good quality always creates a good impression. Proper colour of the paper must be selected (generally white colour). A quality paper is useful to keep records for long period.

(b) Typing:
The typed letters are more attractive and legible than handwritten letters. However, the typing must be neat, accurate, and clean. Spelling must be checked for its correctness. Electronic typewriters or computers with printers must be used.

(c) Margin:
The proper margin on both sides gives the letter a presentable appearance. The size of the margin depends upon the size of the paper. It is also necessary to leave space at the bottom of the letter.

(d) Spacing:
Proper and enough space between the lines should be left. Spacing should be uniform. Proper spacing between the paragraphs increases the attractiveness of letter writing.

(e) Letterhead:
The design of the letterhead should be simple but attractive and impressive. It should include all necessary details like address, telephone number, E-mail, Fax No. etc. The letterhead has advertising value. It introduces the sender to the receiver of the letter.

(f) Folding:
Though the folding of a letter is a simple matter but not an unimportant one. Folds should be minimum. The number of folds depends on the size of paper used for writing a letter and the size of the envelope used for sending it.

(g) Envelope:
The envelope used for sending a letter should be of the proper size, color, and quality. A right-sized envelope should be used. In simple words, the envelope should neither be too small nor too big. If a window envelope is used, to save time and labour then the letter must be folded in such a way that the inside address shall appear exactly on the window (of the envelope).
It Must Be Noted That – “There Are No Exact Rules for Physical Appearance of A Business Letter. It Depends on Circumstances.”

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 10 Correspondence with Directors Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 10 Correspondence with Directors

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Secretary is ____________ to Directors.
(a) owner
(b) servant
(c) member
Answer:
(b) servant

Question 2.
Directors are the ____________
(a) owners
(b) representative of shareholders
(c) creditors of the company
Answer:
(b) representative of shareholders

Question 3.
The provision regarding qualification shares of a director is contained in the ____________
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
Answer:
(a) Articles of Association

1B. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A person is elected by the shareholders of a company.
Answer:
Director

Question 2.
Maximum information in minimum words.
Answer:
Brevity

1C. Complete the sentences.

Question 1.
The Directors are ____________ of shareholders.
Answer:
representative

Question 2.
The Directors are responsible for making ____________
Answer:
decision, plans and policies

Question 3.
The report prepared by the directors at the end of every Financial Year is called ____________
Answer:
Directors report

Question 4.
The gap between two consecutive Board meetings should not be more than ____________
Answer:
120 days

1D. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) ………………………	The link between members and the directors
(2) Brevity	……………………………..

(Secretary, Concise and compact information)
Answer:

Group ‘A’	Group ‘B’
(1) Secretary	The link between members and the directors
(2) Brevity	Concise and compact information

1E. Correct the underlined word and rewrite the following sentences.

Question 1.
Brevity means not using harsh words while writing a letter.
Answer:
Politeness means not using harsh words while writing a letter.

Question 2.
The First Board meeting should be held with 60 days from the date of its incorporation.
Answer:
The First Board meeting should be held with 30 days from the date of its incorporation.

Question 3.
Accuracy means avoiding unnecessary details and irrelevant information in a letter.
Answer:
Brevity means avoiding unnecessary details and irrelevant information in a letter.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions

Question 1.
What is solution?
Answer:
The solution is a homogeneous mixture of two or more components or pure substances. When the size of particles of the components is of the order of 10^-10 m, then the solution is called a true solution. E.g. An aqueous solution of sugar.

Question 2.
Explain : (1) Homogeneous solution
(2) Heterogeneous solution.
Answer:
(1) Homogeneous solution : A solution in which solute and solvent form uniform homogeneous one phase due to attraction between their molecules/particles is called homogeneous solution.
E.g. A solution of NaCl or sugar.

(2) Heterogeneous solution : A solution consisting of two or more phases is called a heterogeneous solution. E.g. A colloidal solution of starch.

Question 3.
Define : (1) Solvent (2) Solute.
Answer:
(1) Solvent : The component in which solution formation takes place and which constitutes larger proportion of a solution is called solvent. For example, in an aqueous solution of sugar, water is the solvent.

(2) Solute : In a solution the component which dissolves and constitutes smaller proportion of a solution is called a solute. For example, in a sugar solution, sugar is the solute.

Question 4.
What are the different types of solutions?
Answer:
A solution consists of a solvent and a solute. Since the physical states of a solvent and a solute may be gaseous, liquid or a solid, there are nine types of solutions.

Question 5.
Mention the solvent and solute in the following :
(1) Smoke
(2) Moisture
(3) Alloy
(4) Soda water.
Answer:

Substance	Solvent	Solute
1. Smoke	Gas	Solid
2. Moisture	Gas	Liquid
3. Alloy	Solid	Solid
4. Soda water	Liquid	Gas

Question 6.
What is a saturated solution?
Answer:
A solution which contains the maximum amount of dissolved solute and further the solute can’t be dissolved is called a saturated solution.

There exists a dynamic equilibrium which is represented as,

Question 7.
What is a supersaturated solution?
Answer:
A solution containing a solute more than that required to form a saturated solution at equilibrium is called a supersaturated solution.

When a tiny crystal of a solute is added to supersaturated solution, the excess solute separates out and forms saturated solution.

Question 8.
What is solubility of a solute ?
Answer:
Solubility : It is defined as amount of a solute present per unit volume in its saturated solution at a specific temperature.
It is expressed in mol L^-1 or mol dm^-3.

Question 9.
Explain the factors on which the solubility of a substance (solute) depends.
Answer:
The extent of dissolution of a substance (solute) depends upon the following factors :
(1) Nature of a solute : A solute may be crystalline, amorphous, ionic or covalent. Hence accordingly its tendency to dissolve changes. The substances having similar intermolecular forces tend to dissolve in each other.

(2) Nature of solvents : Solvents are classified as polar and nonpolar. Polar solutes dissolve in polar solvents. For example, ionic compounds dissolve in polar solvent like water. A solvent other than water is called a nonaqueous solvent. For example, C₆H₆, CCl₄, etc. Solutions in these solvents are called nonaqueous solutions.
The solvent may be a gas, a liquid or a solid.

(3) Amount of a solvent : More amount of a solvent, will dissolve more quantity of the solute.

(4) Temperature : Depending on the nature of a solvent and a solute the solubility changes with termperature. The effect depends on the heat of solution, hydration energy, etc. Generally as the temperature increases, solubility of solid increases and that of gases decreases.

(5) Pressure :

• Pressure has no effect on the solubilities of solids and liquids since they are incompressible.
• The effect of pressure is important only for solutions which involve gases as solutes. With the increase in pressure and decrease in temperature, the solubility of gases increases.

Question 10.
Explain with the help of Le Chatelier’s principle the effect of temperature on solubility.
Answer:

1. The effect of temperature on solubility depends on enthalpy of solution.
2. For example, dissolution of KCl in water is an endothermic process since heat is absorbed during dissolution. In according to Le Chatelier’s principle by increasing temperature the solubility of KCl increases.
3. Dissolution of CaCl₂, Li₂SO₄, H₂O in water is an exothermic process since heat is evolved during dissolution. In this, according to Le Chatelier’s principle by increasing the temperature the solubility decreases.

Question 11.
Explain the solubility of gases in liquids.
Answer:

• Gases are soluble in water and other liquids and their solubility depends upon the nature of the gas.
• Non-polar gases like O₂, have less solubility in polar solvents.
• Polar gases like CO₂, NH₃, HCl, etc. are more soluble in polar solvent like water. CO₂ forms H₂CO₃, while NH₃ forms NH₄OH in aqueous solutions.
• The solubility of gases in liquids increases with the increase in pressure and the decrease in temperature.

Question 12.
Why does the solubility of a gas decrease with increase in temperature ?
OR
How does solubility of a gas in water varies with temperature ?
Answer:

• The gases are soluble in water and other liquids.
• According to Charles’ law, the volume of a given mass of a gas increases with the increase in temperature at constant pressure.
• Hence, the volume of the dissolved gas increases with the increase in temperature.
• This enormous increase in volume of the gases cannot be accommodated by the solvent molecules, hence excess of the gases escape out in the form of bubbles.

Therefore, the solubility of gases in liquids decreases with temperature.

Question 13.
Explain the effect of pressure on the solubility of the gases.
OR
State and explain Henry’s law.
Answer:
(1) Since the gases are compressible, their solubility in the liquids is influenced by external pressure of the gas. The solubility of gases increases with the increase in pressure.

(2) Henry’s law : It states that the solubility of a gas. in a liquid at constant temperature is proportional to the pressure of the gas above the solution.
(i) If S is the solubility of a gas in mol dm^-3 at a pressure P and constant temperature then by Henry’s law,
S ∝ P or S = K_H × P
where K_H is called Henry’s law constant.
(ii) If P = 1 atm, then S = K_H.
(iii) If several gases are present, then the solubility of any gas in the mixture is proportional to its partial pressure at given temperature.

(3) Illustration of Henry’s law : In case of aerated or carbonated drink beverage, the bottle is filled by dissolving CO₂ gas at high pressure and then sealed.
Above the liquid surface there is air and undissolved CO₂. Due to high pressure, the amount of dissolved CO₂ is large.

When the cap of the aerated bottle is removed, the pressure on the solution is lowered, hence excess of CO₂ and air escape out in the form of effervescence. Thus by decreasing the pressure, solubility of CO₂ is decreased.

Question 14.
What are the exceptions to Henry’s law? Why ?
Answer:
(1) The gases like NH₃ and CO₂ do not obey Henry’s law.
(2) This is because, these gases react with water,
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \longrightarrow \mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{OH}_{(\text {aq })}^{-}\)
CO_2(g) + H₂O(l) → H₂CO_3(aq)
(3) Due to reactions of the gases like NH₃, CO_2(g), they have higher solubilities than expected by Henry’s law.

Question 15.
Oxygen gas is slightly soluble in water but it is highly soluble in blood. Explain.
Answer:
The vital constituent of blood, namely haemoglobin reacts with oxygen increasing the solubility of oxygen.
Haemoglobin + 4O_2(g) → Haemoglobin O₈
This oxygenated blood is circulated to the various parts of body, for the supply of oxygen.

Question 16.
Obtain the units of Henry’s law constant.
Answer:
By Henry’s law, S = K_H × P, where S is solubility of the gas in mol dm^-3, P is the pressure of the gas in atmosphere (or in bar) and K_H is Henry’s law constant.
∴ K_H = \(\frac{S^{\left(\mathrm{mol} \mathrm{dm}^{-3}\right)}}{P_{(\mathrm{atm})}}\) mol dm^-3 atm^-1 or mol dm^-3 bar^-1
Hence the units of Henry’s law constant K_H are mol dm^-3atm^-1.

Solved Examples 2.4

Question 17.
Solve the following :
(1) For a gas, the Henry’s law constant is 1.25 × 10^-3 mol dm^-3 atm^-1 at 25 °C. Calculate the solubility of the given gas at 2.5 atm and 25 °C.
Solution :
Given : Henry’s law constant = K_H
= 1.25 × 10^-3 mol dm^-3 atm^-1
Pressure of the gas = P = 2.5 atm
Solubility of the gas = S = ?
By Henry’s law
S = K_H × P
= 1.25 × 10^-3 mol dm^-3 atm^-1 × 2.5 atm
= 3.125 × 10^-3 mol dm^-3
Ans. Solubility of gas = 3.125 × 10^-3 mol dm^-3

(2) The solubility of dissolved oxygen to 27 °C is 2.6 × 10^-3 mol dm^-3 at 2 atm. Find its solubility at 8.4 atm and 27 °C.
Solution :
Given : Solubility of O₂
= S₁ = 2.6 × 10^-3 mol dm^-3
Initial pressure of O₂ = P₁ = 2 atm
Final pressure of O₂ = P₂ = 8.4 atm
Solubility of O₂ = S₂ = ?
(i) By Henry’s law,
S₁ = K_H × P₁
∴ Henry’s law constant K_H is,
K_H = \(\frac{S_{1}}{P_{1}}=\frac{2.6 \times 10^{-3}}{2}\)
= 1.3 × 10^-3 mol dm^-3 atm^-1
(ii) Now, S₂ = K_H × P₂ = 1.3 × 10^-3 × 8.4
= 10.92 × 10^-3
= 1.092 × 10^-2 mol dm^-3
Ans. Solubility of O₂ = 1.092 × 10^-2 mol dm^-3

(3) Henry’s law constant for the solubility of methane in benzene is 4.27 × 10^-5 mm^-1 Hg mold dm^-3 at constant temperature. Calculate the solubility of methane at 760 mm Hg pressure at same temperature.
Solution :
Given : Henry’s law constant = K_H
= 4.27 × 10^-5mm^-1 Hg mol dm^-3
Pressure of the gas = P = 760 mm Hg
K_H = 4.27 × 10^-5 mm^-1 mol dm^-3
= 4.27 × 10^-5 × 760 atm^-1 mol dm^-3
= 3245 × 10^-5 atm^-1 mol dm^-3
= 3.245 × 10^-2 atm^-1 mol dm^-3
P = 760 mm = \(\frac{760}{760}\) atm = 1 atm
By Henry’s law,
S = K_H × P = 3.245 × 10^-2 atm^-1 mol dm^-3 × 1 atm
= 3.245 × 10^-2 mol dm^-3
Ans. Solubility of methane = 3.245 × 10^-2 mol dm^-3

(4) The solubility of ethane at 25 °C is 0.92 × 10^-3 g dm^-3 at 1000 mm Hg pressure. Calculate Henry’s law constant.
Solution :
Given : Solubility of ethane
= S = 0.92 × 10^-5 g dm^-3
Pressure of ethane = P = 1000 mm Hg
Molar mass of ethane (C₂H₆) = 30 g mol^-1
Henry’s law constant = K_H = ?
S = 0.92 × 10^-3 g dm^-3
= \(\frac{0.92}{30}\) × 10^-3
= 3.067 × 10^-5 mol dm^-3
P = 1000 mm = \(\frac{1000}{760}\)atm = 1.316 atm
By Henry’s law,
S = K_H × P
∴ K_H = \(\frac{S}{P}=\frac{3.067 \times 10^{-5}}{1.316}\)
= 2.33 × 10^-5 mol dm^-3 atm^-1
Ans. Henry’s law constant = K_H
= 2.33 × 10^-5 mol dm^-3 atm^-1

(5) The solubility of nitrogen at 30 °C is 2.5 × 10^-3 g dm^-3 at 760 mm pressure. What will be its solubility in mol dm^-3 at 20,000 mm and same temperature?
Solution :
Given : Initial solubility of N₂ = S₁ =2.5 × 10^-3 g dm^-3
Initial pressure = P₁ = 760 mm
Final pressure = P₂ = 20,000 mm
Final solubility = S₂ in mol dm^-3 = ?
Molar mass of N₂ gas = 28 g mol^-1
S₁ = 2.33 × 10^-5 mol dm^-3
= \(\frac{2.5 \times 10^{-3}}{28}\) mol dm^-3
= 8.93 × 10^-5 mol dm^-3
P₁ = \(\frac{760}{760}\) = 1 atm
P₂ = \(\frac{20,000}{760}\) = 26.32 atm
By Henry’s law,
S₁ = K_H × P₁
∴ K_H = \(\frac{S_{1}}{P_{1}}=\frac{8.93 \times 10^{-5}}{1}\)
= 8.93 × 10^-5 mol dm^-3 atm^-1
S₂ = K_H × P₂ = 8.93 × 10^-5 × 26.32
= 2.35 × 10^-3 mol dm^-3
Ans. Solubility of N₂ gas
= 2.35 × 10^-3 mol dm^-3.

[Alternative method:
S₁ = K_HP₁ and S₂ = K_HP₂
∴ \(\frac{S_{2}}{S_{1}}=\frac{K_{\mathrm{H}} P_{2}}{K_{\mathrm{H}} P_{1}}=\frac{P_{2}}{P_{1}}\)
∴ S₂ = S₁ × \(\frac{P_{2}}{P_{1}}\) = 8.93 × 10^-5 × \(\frac{20,000}{760}\)
= 2.35 × 10^-3 mol dm^-3]

Question 18.
Marine life like fish prefers to stay at lower level in water. Explain.
OR
Explain, why do aquatic animals prefer to stay at lower level of water during summer?
Answer:

1. The solubility of oxygen gas decreases with the increase in temperature.
2. In sea or lake water, the temperature of upper level is higher than the lower level.
3. Therefore the dissolved oxygen content in water is more at lower level than at higher level required for the marine life.

Hence marine life like fish prefers to stay at lower level than upper level of water.

Question 19.
State and explain Raoult’s law.
Answer:
Statement of Raoult’s law : The law states that, at constant temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of vapour pressure of the pure component and the mole fraction of that component in the solution.

Let P₀ and P be the respective vapour pressures of a pure volatile component and a solution. If x₁ is the mole fraction of a solvent then by Raoult’s law,
P = x₁ × P₀.

Explanation : Consider a solution containing two volatile components A and B having mole fractions x₁ and x₂ respectively.
Let \(P_{1}^{0}\) and \(P_{2}^{0}\) be the vapour pressures of pure components (or liquids) A and B respectively.
Then by Raoult’s law, vapour pressure of component A = P₁ = X₁ × \(P_{1}^{0}\), vapour pressure of component B = P₂ = x₂ × \(P_{2}^{0}\).
Here P₁ and P₂ represent partial vapour pressures of the two liquid components in the solution.
Hence the total vapour pressure, P_T of the solution will be,
P_T = P₁ + P₂
∴ P_T = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
∵ x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P_T = (1 – x₂)\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= \(P_{1}^{0}\) – x₂\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
With the help of above equation, the vapour pressures of solutions having different concentrations (or mole fractions) can be calculated.

Question 20.
Explain the variation of vapour pressure with mole fraction of a solute in a liquid mixture.
Answer:
Consider a liquid mixture of two liquid components A and B having vapour pressures \(P_{1}^{0}\) and \(P_{2}^{0}\) and mole fractions x₁ and x₂ respectively.
By Raoult’s law, the vapour pressures P₁ and P₂ are,
P₁ = x₁\(P_{1}^{0}\) and P₂ = x₂\(P_{2}^{0}\)
The vapour pressure of the solution is,
P_T = P₁ +P₂ = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)

∴ P_T = (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
The plot of P_T versus x₂ is a straight line. The plots of P₁ versus x₁, and P₂ versus x₂ are straight lines passing through the origin.
When x₁ = 1, x₂ = 0, P_T = \(P_{1}^{0}\) and when x₁ = 0, x₂ = 1, P_T = \(P_{2}^{0}\) as shown by lines I and II in Fig. 2.2

Question 21.
Explain the composition of vapour phase above a liquid mixture.
Answer:
Consider a liquid mixture of two liquids A and B having vapour pressures \(P_{\mathrm{A}}^{0}\) and \(P_{\mathrm{B}}^{0}\) and mole fractions x₁ and x₂ respectively in the liquid phase.
By Raoult’s law, the vapour pressures of two liquids will be,
P_A = x₁\(P_{\mathrm{A}}^{0}\) and P_B = x₂\(P_{\mathrm{B}}^{0}\)
The total vapour pressure of this liquid mixture is,
P_T = P_A + P_B
P_T = x₁\(P_{\mathrm{A}}^{0}\) + x₂\(P_{\mathrm{B}}^{0}\)
The vapour above liquid surface contains A and B. If y₁ and y₂ are the mole fractions of A and B components respectively in the vapour phase, then by Dalton’s law of partial pressures,
P_A = y₁P_T and P_B = y₂P_T
and total vapour pressure is,
P_T = y₁P_T + y₂P_T.

Question 22.
What are ideal and nonideal solutions ?
Answer:

• Ideal solutions : These are solutions which obey Raoult’s law over an entire range of concentrations at constant temperature.
• Nonideal solutions : These are solutions which do not obey Raoult’s law over the entire range of concentrations.

Question 23.
What are the characteristics of ideal solutions ?
Answer:

• The ideal solutions obey Raoult’s law over entire range of concentrations at constant temperature.
• In the formation of an ideal solution, heat is neither evolved nor absorbed and enthalpy change for mixing is zero, i.e. Δ_mix H = 0.
• In the formation of an ideal solution, there is no volume change on mixing two liquid components and the volume of solution is equal to the sum of volumes of two liquid components. Δ_mix V = 0
• In the ideal solution, solvent-solvent, solute-solute and solvent-solute interactions are comparable.
• The vapour pressure of an ideal solution lies between vapour pressures of two pure components.

Question 24.
Give an example of an ideal solution.
Answer:
A liquid mixture of benzene and toluene which have nearly identical physical properties and inter molecular forces forms an ideal solution.

Question 25.
What are the characteristics of nonideal solutions.
Answer:

• Nonideal solutions do not obey Raoult’s law over the entire range of concentrations.
• The vapour pressures of these solutions may be higher or lower than ideal solutions.
• These solutions exhibit two types of deviations from Raoult’s law namely (a) positive deviation and (b) negative deviation.
• These solutions give azeotropic mixtures.

Question 26.
Explain solutions with positive deviations from Raoult’s law.
Answer:
(i) A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult’s law or higher than those of pure components is called a nonideal solution with positive deviation.

(ii) In these solutions, solute-solvent intermolecular attractions are weaker than those between solvent-solvent and solute-solute interactions.
(iii) For example, solutions of acetone and ethanol, carbon disulphide and acetone, etc.

Question 27.
Explain solutions with negative deviations from Raoult’s law.
Answer:
(1) A solution or a liquid mixture which has lower vapour pressure than theoretically calculated by Raoult’s law or lower than those of pure components is called a nonideal solution with negative deviation.

(2) In these solutions, the intermolecular interactions between solvent and solute molecules are stronger than solvent-solvent or solute-solute interactions.
(3) For example, solutions of phenol and aniline, chloroform and acetone, etc.

Solved Examples 2.5

Question 28.
Solve the following :

(1) The vapour pressures of two liquids A and B are 400 mm Hg and 600 mm Hg respectively at 47 °C. A solution is prepared by dissolving 10 g of A of molar mass 60 g mol^-1 in 80 g of B of molar mass 40 g mol^-1. Find the vapour pressure of the solution.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg.
W_A = 10 g and W_B = 80 g.
M_A = 60 g mol^-1; M_B = 40 g mol^-1, P_soln = ?
n_A = \(\frac{W_{\mathrm{A}}}{M_{\mathrm{A}}}=\frac{10}{60}\) = 0.1667 mol
n_B = \(\frac{W_{\mathrm{B}}}{M_{\mathrm{B}}}=\frac{80}{40}\) = 2 mol
Total number of moles = n = n_A + n_B
= 0.1667 + 2
= 2.1667 mol

= 0.07693 × 400 + 0.9230 × 600
= 30.772 + 553.8
= 584.572 mm Hg
Ans. Vapour pressure of solution
= P_soln = 584.572 mm Hg

(2) 2.5 mol of a liquid A is mixed with 4.5 mol of liquid B at 25 °C. If the vapour pressures of A and B at 25 °C are 160 mm Hg and 230 mm Hg respectively, calculate the vapour pressure of the liquid mixture.
Solution :
Given : n_A = 2.5 mol; n_B = 4.5 mol,
\(P_{\mathrm{A}}^{0}\) = 160 mm Hg; \(P_{\mathrm{B}}^{0}\) = 230 mm Hg, P_soln = ?
Total number of moles = n = n_A + n_B
= 2.5 + 4.5
= 7.0 mol
Mole fraction of A = x_A = \(\frac{n_{\mathrm{A}}}{n}=\frac{2.5}{7}\) = 0.3571
Mole fraction of B = 1 – x_A = 1 – 0.3571
= 0.6429
\(P_{\text {soln }}=x_{\mathrm{A}} P_{\mathrm{A}}^{0}+x_{\mathrm{B}} P_{\mathrm{B}}^{0}\)
= 0.3571 × 160 + 0.6429 × 230
= 57.136 + 147.9
= 205 mm Hg
Ans. P_soln = 205 mm Hg

(3) The vapour pressures of pure liquids A and B are 400 mm Hg and 650 mm Hg respectively at 330 K. Find the composition of liquid and vapour if total vapour pressure of solution is 600 mm Hg.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg,
P_T = 600 mm Hg, T = 330 K; x_A = ? x_B = ?, y₁ = ? y₂ = ?
(x is mole fraction in liquid phase while y is mole fraction in vapour phase.)
P_T = (\(P_{\mathrm{A}}^{0}\) – \(P_{\mathrm{B}}^{0}\))x_B + \(P_{\mathrm{A}}^{0}\)
600 = (650 – 400)x_B + 400
= 250x_B + 400
∴ x_B = \(\frac{600-400}{250}\) = 0.8
∵ x_A + x_B = 1
∴ x_A = 1 – x_B = 1 – 0.8 = 0.2
The composition of A and B in liquid mixture is, x_A = 0.2 and x_B = 0.8.
If P₁ and P₂ are vapour pressures (or partial pressures) of A and B in vapour phase then by Raoult’s law,
P₁ = x_A × \(P_{\mathrm{A}}^{0}\) = 0.2 × 400 = 80 mm Hg
P₂ = x_B × \(P_{\mathrm{B}}^{0}\) = 0.8 × 650 = 520 mm Hg
If y₁ and y₂ are mole fractions of A and B respectively in vapour phase then, by Dalton’s law,
P₁ = y₁P_T

(or y₂ = 1 – y₁ = 1 – 0.1333 = 0.8667)
Ans. Composition of liquid : x_A = 0.2 and x_B = 0.8
Composition of vapour: y_A = 0.1333 and y_B = 0.8667

(4) A mixture of two liquids A and B have vapour pressures 3.4 × 10⁴ Nm^-2 and 5.2 × 10 Nm^-2. If the mole fractions of A is 0.85, find the vapour pressure of the solution.
Solution:
Given : Vapour pressure of pure liquid A
= \(P_{\mathrm{A}}^{0}\) = 3.4 × 10⁴ Nm^-2
Vapour pressure of pure liquid B
= \(P_{\mathrm{B}}^{0}\) = 5.2 × 10⁴ Nm^-2
Mole fraction of A = x_A = 0.85
Mole fraction of B = x_B = 1 – x_A
= 1 – 0.85
= 0.15
The vapour solution is given by
\(P_{\text {soln }}=X_{A} P_{A}^{0}+X_{B} P_{B}^{0}\)
= 0.85 × 3.4 × 10 + 0.15 × 5.2 × 10⁴
=2.89 × 10⁴ + 0.78 × 10⁴
=(2.89 + 0.78) × 10⁴
P_soln = 3.67 × 10⁴ Nm^-2
Ans. Vapour pressure of a solution = 3.67 × 10⁴ Nm^-2

Question 29.
Define the term colligative property. Give examples.
Answer:
(1) Colligative Property : The property of a solution which depends on the total number of particles of the solute (molecules, ions) present in the solution and does not depend on the nature or chemical composition of solute particles is called colligative property of the solution.

(2) Examples of colligative properties : (a) lowering or relative lowering of vapour pressure of a solution (b) elevation in the boiling point (c) depression in the freezing point (d) osmotic pressure.

Question 30.
Explain and define the term vapour pressure of a liquid.
OR
What is vapour pressure of a liquid?
Answer:

(1) If a volatile liquid is placed in an open vessel, the liquid molecules have a tendency to escape in a gaseous state forming vapour and diffuse into surroundings. Hence evaporation takes place continuously, and no equilibrium is attained.

(2) If a liquid is placed in a closed vessel, the vapour molecules get accumulated on its surface. These vapour molecules are in continuous random motion. They collide with each other, with walls of a container and surface of the liquid, and return to the liquid state. This reverse phenomenon is called condensation.

(3) After some time, rates of evaporation and condensation become equal and an equilibrium is established between liquid and vapour phases. At this stage the vapour exerts a constant pressure called vapour pressure on liquid surface at constant temperature.

(4) Vapour pressure : The pressure exerted by the vapour of a liquid (or solid) when it is in equilibrium with the liquid (or solid) phase at a constant temperature is called the vapour pressure of the liquid (or solid).

(5) The vapour pressure of a liquid increases with the increase in temperature.

Question 31.
Explain the following terms :
(1) Relative vapour pressure of a solution
(2) Lowering of vapour pressure of a solution
(3) Relative lowering of vapour pressure.
Answer:
(1) Relative vapour pressure of a solution : If Po is the vapour pressure of a pure liquid (solvent) and P is the vapour pressure of a solution after adding a nonvolatile solute, then, relative vapour pressure = \(\frac{P}{P_{0}}\).

(2) Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.
If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

(3) Relative lowering of vapour pressure : If P₀ and P are the respective vapour pressures of a pure liquid (solvent) and the solution containing a non-volatile solute then P < P₀. Hence, P₀ – P represents the lowering of the vapour pressure due to addition of a nonvolatile solute.
∴ Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}=\frac{\Delta P}{P_{0}}\)

Question 32.
State and explain Raoult’s law for solutions of nonvolatile solutes.
Answer:
(a) Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

(b) Explanation : Let P0 and P be the vapour pressures of a pure solvent and a solution respectively. If x₁ is the mole fraction of the solvent then Raoult’s law can be represented as,
P = x₁P₀
For a binary solution containing one solute, if x₁ and x₂ are mole fractions of a solvent and a solute respectively then,
x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P = x₁P₀
= (1 – x₂) P₀
= P₀ – x₂P₀
= P₀ – P = x₂P₀
∴ P₀ – P = x₂P₀
∴ x₂ = \(\frac{P_{0}-P}{P_{0}}\)
P₀ – P = ΔP is the lowering of vapour pressure
∴ x₂ = \(\frac{\Delta P}{P_{0}}\)
In the equation, P₀ – P/P₀ is called relative lowering of vapour pressure.
Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.

Question 33.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
Consider a binary solution containing a nonvolatile solute. If P₀ and P are vapour pressures of a pure solvent and the solution respectively then, Lowering of vapour pressure = ΔP = P₀ – P
Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law,
\(\frac{P_{0}-P}{P_{0}}=x_{2}\)
where x₂ is mole fraction of the solute. Therefore the relative lowering of vapour pressure is a colligative property.

Question 34.
Explain the variation of vapour pressure with mole fraction of a solvent in solution.
OR
Explain the variation of vapour pressure with the concentration of a solution.
Answer:
The vapour pressure of a pure solvent decreases when a nonvolatile solute is dissolved in it. Consider a pure solvent with vapour pressure P₀ and mole fraction x₁. For a solution containing a non-volatile solute, if x₁ and x₂ are the mole fractions of a solvent and a solute respectively, then x₁ + x₂ = 1 and x₁ < 1.

By Raoult’s law, P = x₁ × P₀. As the mole fraction of a solvent in the solution increases the vapour pressure increases as shown in the above figure 2.5. When x₁ becomes equal to 1, the vapour pressure becomes P₀, i.e. the vapour pressure of a pure solvent.

If at any mole fraction of a solvent, the vapour of the solution is P, then the lowering of vapour pressure will be, ΔP = P₀ – P.

Question 35.
Derive a relation between relative lowering of vapour pressure and molar mass of non-volatile solute.
Answer:
Consider a solution in which W₁ gram of a solvent of molar mass (or molecular weight) M₂ contains W₂ gram of a solute of molar mass M₂. Then
number of moles of a solvent = n₁ = \(\frac{W_{1}}{M_{1}}\)
Number of moles of a solute = n₂ = \(\frac{W_{2}}{M_{2}}\)
∴ Total number of moles = n = n₁ + n₂
Mole fraction of the solvent = x₁ = \(\frac{n_{1}}{n}\)
Mole fraction of the solute = x₂ = \(\frac{n_{2}}{n}\)

In the case of an ideal solution which is a dilute solution, the concentration and the number of moles of the solute are very low, i.e. n₂ << n₁

If P₀ and P are the vapour pressures of a pure solvent and a solution respectively, then relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law

Hence by measuring the vapour pressure of a pure solvent and a solution, the molar mass of the dissolved nonvolatile substances can be determined.

Solved Examples 2.6 – 2.7

Question 36.
Solve the following :

(1) The vapour pressure of a pure liquid at 298 K is 4 × 10⁴ Nm^-2. When a nonvolatile solute is dissolved the vapour pressure becomes 3.65 × 10⁴ Nm^-2. Calculate (A) relative vapour pressure, (B) lowering of vapour pressure and (C) relative lowering of vapour pressure.
Solution :
Given : P₀ = 4 × 10⁴ Nm^-2
P = 3.65 × 10⁴ Nm^-2
(A) Relative vapour pressure = \(\frac{P}{P_{0}}\)
= \(\frac{3.65 \times 10^{4}}{4 \times 10^{4}}\)
= 0.9125

(B) Lowering of vapour pressure = ΔP = P₀ – P
= 4 × 10⁴ – 3.65 × 10⁴
= (4 – 3.65) × 10⁴
= 0.35 × 10⁴ Nm^-2 = 3.5 × 10³ Nm^-2

(C) Relative lowering of vapour pressure is given by,

Ans. (A) 0.9125 (B) 3.5 × 10³ Nm^-2 (C) 0.0875

(2) A pure liquid has vapour pressure 5.2 × 10⁴ Pa at 298 K. When a solute is dissolved, the mole fraction of it is 0.02 in the solution. Find the vapour pressure of the solution.
Solution :
Given : P₀ = 5.2 × 10⁴ Pa
x₂ = 0.02
P = ?

Ans. Vapour pressure of the solution
= 5.096 × 10⁴ Pa

(3) The vapour pressure of pure benzene is 640 mm of Hg. 2.175 × 10^-3 kg of nonvolatile solute is added to 39 gram of benzene, the vapour pressure of solution is 600 mm of Hg. Calculate molar mass of solute (C = 12, H = 1).
Solution :
Given : P₀ = 640 mm Hg
W₁ = 39 g benzene = 39 × 10^-3 kg
W₂ = 2.175 × 10^-3 kg
P = 600 mm Hg
M₁ = 78 × 10^-3 kg, M₂ = ?

Ans. Molar mass of a solute = 69.6 × 10^-3 kg mol^-1

(4) In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.
Solution :
Given : Mass of a solute (mannitol)
= W₂ = 18.04 g
Mass of a solvent (water) = W₁ = 100 g
Vapour pressure of a solvent (water)
= P₀
= 17.535 mm Hg
Lowering of vapour pressure = ΔP = 0.309 mm Hg
Molar mass of H₂O = M₁ = 18 g mol^-1
Molar mass of solute (mannitol) = M₂ = ?

Ans. Molar mass of mannitol = 184.3 g mol^-1

(5) The vapour pressure of 2.1% solution of a non-electrolyte in water at 100 °C is 755 mm Hg. Calculate the molar mass of the solute.
Solution :
Given : At 100 °C, vapour pressure of water = P₀ = 760 mm Hg
Vapour pressure of the solution = P = 755 mm Hg
Since the solution is 2.1 % by mass,
Mass of solute (nonelectrolyte) = W₂ = 2.1 g
Mass of water = W₁ = 100 – 2.1 = 97.9 g
Molar mass of water = M₁ = 18 g mol^-1
Molar mass of nonelectrolyte = M₂ = ?

Ans. Molar mass of nonelectrolyte = 58.69 g mol^-1

(6) Calculate the mass of a nonvolatile solute (molar mass 40 × 10^-3 kg/mol) which is dissolved in 114 × 10^-3 kg octane to reduce its vapour pressure to 80%.
Solution :
Given : Molar mass of solute = M₂ =40 × 10^-3 kg mol^-1
Mass of solvent (octane) = W₁ = 114 × 10^-3 kg
If vapour pressure of octane = P₀ = 100
Vapour pressure of solution = P = 80
Molar mass of octane (C₈H₁₈) = 114 × 10^-3 kg mol^-1
Mass of solute = W₂ = ?
In this solution, since the vapour pressure is decreased by greater extent, 100 – 80 = 20% the solution must be concentrated and number of moles of solute must be large so that n₁ + n₂ ≠ n₁
Hence Raoult’s law must be represented as,

∴ Mass of solute dissolved = W
= moles × molar mass = 0.25 × 40 × 10^-3
= 0.01 kg
Ans. Mass of solute dissolved = 0.01 kg

(7) The vapour pressure of water is 16.8 mm Hg at a certain temperature. If the vapour pressure of the solution is 16.78 mm, find the molality of the solution.
Solution :
Given : P₀ = 16.8 mm Hg; P = 16.78 mm Hg
Molar mass of water = M₁ = 18 g mol^-1
Molality of the solution = m = ?

Ans. Molality of the solution = 0.66 m

Question 37.
Explain the effect of temperature on the vapour pressure of a liquid.
Answer:
The vapour pressure of a liquid is the pressure of the vapour in equilibrium with the liquid at a given temperature. The evaporation of a liquid requires thermal energy. Hence, as temperature rises, the vapour pressure rises until it becomes equal to the external pressure, generally the atmospheric pressure, 101.3 kNm^-2 (1 atm). This temperature is called the normal boiling point of the liquid.

Question 38.
What is boiling point of liquid?
OR
Define boiling point.
Answer:
The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, i.e., the atmospheric pressure (1 atm), e.g., the boiling point of water at 1 atm is 373 K. m

Question 39.
Explain the elevation in the boiling point of a solution.
Answer:
The elevation in the boiling point of a solution is defined as the difference between the boiling points of the solution and the pure solvents at a given pressure, e.g. If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in boiling point, ΔT_b = T – T₀. It is a colligative property.

Question 40.
What are the units of molal elevation constant?
Answer:
Boiling point elevation, ΔT_b is given by,
ΔT_b = K_b × m
where m is molality in mol kg^-1 and K_b is molal elevation constant or abullioscopic constant.

∴ K_b has units K kg mol^-1 (or °C kg mol^-1)
Therefore, molal elevation constant is the elevation in boiling point produced by 1 molal solution of a nonvolatile solute.

Question 41.
Derive the relation between molar mass of the solute and boiling point elevation.
Answer:
The boiling point elevation, ΔT_b of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_b ∝ m
ΔT_b = K_b m
where K_b is a proportionality constant
If m = 1 molal, then
ΔT_b = K_b
where K_b is called molal elevation constant.
The molality of the solution is given by,
Number of moles of the solute, m = \(\frac{\text { Number of moles of the solute }}{\text { Weight of the solvent in } \mathrm{kg}}\)
Let W₁ = Weight (in gram) of a solvent,
W₂ = Weight (in gram) of a solute
M₂ = Molecular weight of the solute
Then the molality (m) of the solution is given by

If the weights and molecular weight are expressed in kg, then,
\(\Delta T_{\mathrm{b}}=K_{\mathrm{b}} \times \frac{W_{2}}{W_{1} M_{2}}\)

Question 42.
Define molal elevation constant (Ebullioscopic constant)? Does it depend on the nature of a solute ? What are its units ?
Answer:
(1) Molal elevation constant (Ebullioscopic constant) : It is defined as the elevation in boiling point, produced by dissolving one mole of a solute in 1 kg (or 1000 gram) of a solvent (i.e. 1 molal solution).
The elevation in the boiling point,
ΔT_b is given by ΔT_b = K_b × m
where Kb is molal elevation constant and m is molality of the solution.
∴ When m = 1, ΔT_b = K_b
(2) K_b depends only on the nature of the solvent.
(3) K_b does not depend on the nature of the solute.
(4) It does not depend on concentration of the solution.
(5) The units of molal elevation constant are :
(A) K kg mol^-1 and (B) Km^-1.

Solved Examples 2.8

Question 43.
Solve the following :

(1) Calculate the (i) elevation in the boiling point and (ii) the boiling point of 0.05 m aqueous solution of glucose. (K_b = 0.52 Km^-1)
Solution :
Given : Concentration of the solution = m = 0.05 m
Molal elevation constant = K_b = 1.86 K kg mol^-1
Boiling point of pure water = T₀ = 373 K
Elevation in the boiling point = ΔT_b = ?
Boiling point of solution = T_b = ?

(i) ΔT_b = K_b × m
= 0.52 × 0.05
= 0.026K

(ii) The elevation in the boiling point is given by,
ΔT_b = T_b – T₀
∴ Boiling point of a solution,
T_b = T₀ + ΔT_b
= 373 + 0.026
= 373.026 K
Ans. ΔT_b = 0.026 K, T_b = 373.026 K

(2) 0.18 molal aqueous solution of a substance boils at 373.25 K. Calculate the molal elevation constant of water. (Boiling point of water is 373.15 K)
Solution :
Given : Concentration of solution = m = 0.18 m
Boiling point of water = T₀ = 373.15 K
Boiling of the solution = T_b = 373.25
Molal elevation constant = K_b = ?
Elevation in boiling point = ΔT_b
= T_b – T₀
= 373.25 – 373.15
= 0.1 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}=\frac{0.1}{0.18}\) = 0.5556 K kg mol^-1
Ans. Molal elevation constant of water
= 0.5556 K kg mol^-1

(3) The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. [K_b for benzene = 2.53 K kg mol^-1]
Solution :
Given : \(T_{b}^{0}\) = 353.23 K; T_b = 354.11 K
W₁ = 90 g; W2 = 1.8 g; K_b = 2.53 K kg mol^-1
Molar mass = M₂ = ?
ΔT_b = T_b – \(T_{b}^{0}\) = 354.11 – 353.23 = 0.88 K

Ans. Molar mass = M₂ = 57.5 g mol^-1

(4) Boiling point of a solvent is 80.2°C. When 0.419 g of the solute of molar mass 252.4 g mol^-1 is dissolved in 75 g of the above solvent, the boiling point of the solution is found to be 80.256 °C. Find the molal elevation constant.
Solution :
Given : T₀ = (273 + 80.2) K
T_b = 273 + 80.256 (K)
W₁ = 75g
W₂ = 0.419 g
M₂ = 252.4 g mol^-1
K_b = ?
ΔT_b = T_b – T₀
= (273 + 80.256) – (273 + 80.2)
= 0.056 K

Ans. K_b = 2.53 K kg mol^-1

(5) 3.795 g of sulphur is dissolved in 100 g of CS₂. This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? Theboiling point of CS₂ is 319.45 K.
(Given that K_b for CS₂ =2.42 K kg mol^-1 and atomic mass of S = 32.)
Solution :
Given : W₂ = 3.795 g; W₁ = 100 g CS₂
T_b = 319.81 K; T₀ = 319.45 K
K_b = 2.42 K kg mol^-1 M₂ = ?
ΔT_b = T_b – T₀ = 319.81 – 319.45 = 0.36 K

Number of S atoms in molecule = \(\frac{255.1}{32}\)
= 7.972
≅ 8
Ans. Formula of sulphur in CS₂ solution = S₈

(6) Calculate the mass in grams of an impurity of molar mass 100 g mol^-1 which would be required to raise the boiling point of 50 g of chloroform by 0.30°C. (K_b for chloroform = 3.63 K kg mol^-1)
Solution :
Given : M₂ = 100 g mol^-1
W₁ = 50 g chloroform
W₂ = ?
ΔT_b = 0.30°C.
K_b = 3.63 K kg mol^-1

Ans. Weight of impurity = 0.4132 g

(7) A solution containing 0.5126 g of naphthalene (molar mass = 128.17 g mol^-1) in 50.0 g of CCl₄ gives a boiling point elevation of 0.402 °C. While a solution of 0.6216 g of unknown solute in the same mass of the solvent gives a boiling point elevation of 0.647 °C. Find the molar mass of the unknown solute. (K_b for CCl₄ = 5.03 K kg mol^-1 of solvent)
Solution:
Given : For naphthalene solution :
W₂ = 0.5126 g
W₁ = 50 g
M₂ = 128.17 g mol^-1
K_b = 5.03 K kg mol^-1
ΔT_b = 0.402°C
For unknown solute :
W’₁ = 50g
W’₂ = 0.6216 g
ΔT’_b = 0.647°C
M’₂ = ?
For the solution of unknown substance,

Ans. Molecular weight of unknown substance = 96.65 g mol^-1

(8) A solution containing 0.73 g of camphor (molar mass 152 g mol^-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46°C. Calculate the molar mass of the unknown compound.
Solution:
Given : Mass of solute = W₂ = 0.73 g camphor
Mass of solvent = W₁ = 36.8 g
Molar Mass of solute = M₂ = 152 g mol^-1
Boiling point of acetone = T₀ = (273 + 56.3) K
Mass of unknown solute = W’₂ = 0.564 g
Mass of solvent = W’₁ = 36.8 g
Boiling point of solution of camphor = T_b = (273 + 56.55) K
Boiling point solution of unknown compound = T’_b = (273 + 56.46) K
Molar mass of unknown compound = M’₂ = ?

For camphor solution,
∴ ΔT_b = T_b – T₀
= (273 + 56.55) – (273 + 56.3)
= 0.25 K

For a solution of unknown compound
ΔT’_b = ΔT’_b – T₀
= (273 + 56.46) – (273 + 56.3)
= 0.16 K

Ans. Molar mass of the compound = 183.5 g mol^-1

(9) 0.12 molal solution of a substance boils at 373.21°K. Calculate molal elevation constant of the solvent.
Solution :
Given : Concentration of solution = m = 0.12
Boiling point of the solution = T_b = 373.21 K
For solvent (water) T₀ = 373.15 K
Molal elevation constant, K_b = ?
ΔT_b = T_b – T₀
= 373.21 – 373.15
= 0.06 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}\)
= \(\frac{0.06}{0.12}\)
=0.5 K kg mol^-1
Ans. Molal elevation constant = K_b = 0.5 K kg mol^-1

(10) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 5.02 K kg mol^-1)
Solution :
Given : Boiling point of water = T₀ = 273 + 99.63
= 372.63 K
Boiling point of a solution = T_b = 273 + 100 = 373 K
Mass of a solvent (water) = W₁ = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁) = M2
= 342 g mol^-1
K_b = 5.02 K kg mol^-1
Mass of solute (sucrose) = W₂ = ?
ΔT_b = T_b – T₀
= 373 – 372.63
= 0.37 K

Ans. Mass of sucrose required to be added = 12.60 g

(11) A solution of phosphorus prepared by dissolving 0.0175 kg of phosphorus in 0.08 kg of CS₂ has a boiling point 319.87 K. If K_b for CS₂ is 2.4 K kg mol-1 and atomic mass of phosphorus is 31 × 10^-3 kg mol^-1, find the formula of phos-phorus. (Boiling point of CS₂ = 319.45 K)
Solution :
Given : Mass of solvent (CS₂) = W₁ = 0.08 kg
Mass of phosphorus = W₂ = 0.0175 kg
Boiling point of CS₂ = T₀ = 319.45 K
Boiling point of solution = T_b = 319.87 K
Molal elevation constant = K_b = 2.4 K kg mol^-1
Atomic mass of phosphorus = 31 × 10^-3 kg mol^-1
Molecular formula of phosphorus = ?
ΔT_b = T_b – T₀ = 319.87 – 319.45 = 0.42 K

∴ Number of atoms in a molecule of phosphorus = \(\frac{\text { molar mass of phosphorus }}{\text { atomic mass of phosphorus }}\)
= \(\frac {125}{31}\)
= 4.032
≅ 4
Hence the molecular formula of phosphorus is P₄ in CS₂.
Ans. Molecular formula of phosphorus = P₄

(12) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 0.52 K kg mol^-1)
Solution :
Given : Pressure = P = 750 mm Hg
T₀ = 273 + 99.63 = 372.63 K
T_b = 273 + 100 = 373 K
K_b = 0.52 K kg mol^-1
Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 × 10^-3 kg mol^-1
W₁ = 500 g = 0.5 kg
Mass of sucrose to be added = W₂ = ?
ΔT_b = T_b – T₀ = 373 – 372.63 = 0.37 K

Ans. Mass of sucrose to be added = 121.7 × 10^-3 kg
= 121.7 g

(13) 35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.
Solution :
35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O.
Molar mass of water = 18 g mol^-1
Molar mass of ethylene glycol (CH₂OH-CH₂OH) = 62 g mol^-1
Number of moles of water = n₁ = \(\frac{65}{18}\) = 3.611 mol
Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)
= 0.5645 mol
Total moles = n = n₁ + n₂ = 3.611 + 0.5645
= 4.1755 mol
Mole fraction of ethylene glycol = x₂
\(=\frac{n_{2}}{n_{1}}=\frac{0.5645}{4.1755}\) = 0.1352
∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352
= 0.8648
Ans. Mole fraction of water = 0.8648
Mole fraction of ethylene glycol = 0.1352

Question 44.
Define freezing point of a liquid.
Answer:
The freezing point of a liquid is defined as the temperature at which the solid coexists in the equilibrium with the liquid and the vapour pressure of the liquid and the solid are equal.

Question 45.
Explain the depression in the freezing point of a solution.
Answer:
The depression in the freezing point of a solution is defined as the difference between the freezing points of a pure solvent and that of the solution.
If T₀ and T are the respective freezing points of a pure solvent and a solution, then the depression in the freezing point ΔT_f is given by,
ΔT_f = T₀ – T (T < T₀)
The depression in the freezing point (ΔT_f) is a colligative property.

Question 46.
What causes depression in freezing point ?
OR
Explain, freezing point depression as a consequence of vapour pressure lowering.
Answer:
The freezing point of a liquid is the temperature at which the liquid and the solid have the same vapour pressure.

Addition of a nonvolatile solute to a liquid decreases the freezing point, i.e., the freezing point of the solution is less than that of the pure solvent. This is due to the lowering of the vapour pressure of the solvent by the addition of the nonvolatile solute.

When a liquid is cooled from the point A, its vapour pressure decreases and at the point B, it freezes (solidifies).

In case of a solution, since the vapour pressure is lowered, the freezing point decreases. Hence, if the solution is cooled from the point A’, it freezes at lower temperature B’, than the pure liquid. This is also due to separation of solvent molecules due to solute molecules decreasing their intermolecular attraction.

If T₀ and T are the freezing points of a pure solvent and the solution, then, the depression in the freezing point is given by,
ΔT_f = T₀ – T.
This depression ΔT_f depends on the lowering of the vapour pressure (P₀ – P). ΔT_f ∝ (P₀ – P), where P₀ and P are the vapour pressures of the pure liquid and the solution.

Question 47.
What is a relationship between freezing point depression and concentration of solute (or solution)?
Answer:
It is observed experimentally that as the concentration of a solution increases, the freezing point of the solution decreases and hence the depression in the freezing point (ΔT_f) increases.

The depression in the freezing point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Thus,
ΔT_f ∝ m
where m is the molality of the solution.
∴ ΔT_f = Kfm, where Kf is a constant of proportionality. If m = 1 molal,
ΔT_f = K_f. Hence K_f is called the cryoscopic constant or molal depression constant. K_f is characteristic of the solvent.

Question 48.
What are the units of molal depression constant or cryoscopic constant?
Answer:
The freezing point depression, ΔT_f is given by,
ΔT_f = K_f × m
where m is molality in mol kg^-1 and K_f is molal depression constant or cryoscopic constant.

∴ K_f has unit K kg mol^-1 (or °C kg mol^-1)
Therefore cryoscopic constant is the depression in freezing point produced by 1 molal solution of a nonvolatile solute.

Question 49.
Write the formula to determine molar mass of a solute using freezing point depression method.
Answer:
M₂ = \(\frac{K_{\mathrm{f}} \times W_{2} \times 1000}{W_{1} \times \Delta T_{\mathrm{f}}}\)
where
K_f = Molal depression constant
ΔT_f = Depression in freezing point
W₁ = Mass of a solvent
W₂ = Mass of a solute.
M₂ = Molar mass of solute

Question 50.
Define cryoscopic constant (or molal depression constant).
Answer:
Molal depression constant : It is defined as the depression in freezing point, produced by dissolving one mole of a solute in 1 kg (or 1000 g) of a solvent (i.e. 1 molal solution).

Solved Examples 2.9

Question 51.
Solve the following :

(1) 1.35 g of a substance when dissolved in 55 g acetic acid produced a depression of 0.618°C in a freezing point. Calculate the molar mass of the dissolved substance. (K_f = 3.865 K kg mol^-1)
Solution :
Given : Mass of a solvent = W₁ = 55 g
Mass of a solute = W₂ = 1.35 g
Depression in freezing point = ΔT_f = 0.618 °C
Molal depression constant = K_f = 3.865 K kg mol^-1
Molar mass of a solute = M₂ = ?

Ans. Molar mass of the substance = 153.5 g mol^-1

(2) When certain amount of sucrose is dissolved in 1 kg of water, the freezing point of the solution is found to be 272.8 K. If the molecular mass of sucrose is 342 g mol^-1 and K_f for water is 1.86 K kg mol^-1, calculate the amount of sucrose present in the solution.
Solution :
Given : W₁ = Mass of the solvent = 1 kg
T₀ = Freezing point of pure water = 273 K
T_f = Freezing point of the solution = 272.8 K
ΔT_f = T₀ – T_f = Depression in freezing point
= 273 K – 272.8 K
= 0.2 K
M₂ = Molecular mass of the solute
= 342 × 10^-3 kg mol^-1
K_f = Molal freezing point depression constant for water
= 1.86 K kg mol^-1
W₂ = Mass of the solute = ?
ΔT_f = T₀ – T_f

Ans. Mass of sucrose = 36.78 × 10^-3 kg

(3) The freezing point of a pure solvent is 315 K. On addition of 0.5 mole of urea in 1 kg of the solvent, the freezing point decreases by 3 K. Calculate the molal depression constant for the solvent.
Solution :
Given :
m = Molality of urea = 0.5 m
ΔT_f = Depression in the freezing point = 3 K
K_f = Molal depression constant for the solvent = ?
ΔT_f = K_f × m
∴ K_f = \(\frac{\Delta T_{\mathrm{f}}}{m}\)
= \(\frac{3}{0.5}\)
= 6 K kg mol^-1
Ans. Molal depression constant = 6 K kg mol^-1

(4) The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2g of a solute having molecular weight 100 is added to 100 g of benzene. (K_f for benzene = 5.12 K kg mol^-1)
Solution :
Given : K_f for benzene = 5.12 K kg mol^-1
T₀ = Freezing point of the solvent = 278.4 K
W₁ = Mass of the solvent = 100 g = 0.1 kg
W₂ = Mass of the solute = 2g = 2 × 10^-3 kg
M₂ = Molecular mass of the solute = 100 g mol^-1
= 100 × 10^-3 kg mol^-1
T = Freezing point of the solution = ?

Ans. Freezing point of the solution = 277.376 K

(5) 0.635 × 10^-3 kg of a substance of molar mass 190 × 10^-3 kg mol^-1 was dissolved in 30.5 × 10^-3 kg of a solvent. If the depression in the freezing point is 0.62 °C, find the molal depression constant of the solvent.
Solution :
Given : Mass of solvent = W₁ = 30.5 × 10^-3 kg
Mass of solute = W₂ = 0.635 × 10^-3 kg
Depression in freezing point = ΔT_f = 0.62 °C (or K)
Molar mass of the substance = M₂ = 190 × 10^-3 kg mol^-1
Molal depression constant = K_f = ?

Ans. Molal depression constant = K_f
= 5.66 K kg mol^-1

(6) The boiling point of an aqueous solution is 100.18 °C. Find the freezing point of the solution. (Given : K_b = 0.52 K kg mol^-1, K_f = 1.86 K kg mol^-1)
Solution :
Given : T_b = 100.18 °C + 273 = 373.18 K
K_b = 0.52 K kg mol^-1; K_f = 1.86 K kg mol^-1
Boiling point of water = T₀ = 373 K
T_f = ?
ΔT_b = T_b – T₀ = 373.18 – 373 = 0.18 K
If m is the molality of the solution, then
ΔT_b = K_b × m and ΔT_f = K_f × m

Freezing point of water = T₀ = 273 K
ΔT_f = T₀ – T_f
Hence the freezing point of the solution is,
T_f = T₀ – ΔT_f = 273 – 0.6438 = 272.3562 K
OR Freezing point of solution is -0.6438 °C.
Ans. Freezing point of the solution
= 272.3562 K
= -0.6438 °C

(7) 1.0 × 10^-3 kg of urea when dissolved in 0.0985 kg of a solvent, decreases freezing point of the solvent by 0.211 K. 1.6 × 10^-3 kg of another nonelectrolyte solute when dissolved in 0.086 kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
Solution :
Given : Mass of Urea = W₂ = 1 × 10^-3 kg
Mass of solvent = W₁ = 0.0985 kg
Depression in freezing point = ΔT_f = 0.211 K
Molar mass of urea = M₂ = 60 g mol^-1
= 60 × 10^-3 kg mol^-1
Mass of another solute = W’₂ = 1.6 × 10^-3 kg
Mass of solvent = W’₁ = 0.086 kg
Depression in freezing point = ΔT’_f = 0.34 K
Molar mass of another solute = 60 × 10^-3 kg mol^-1

From urea solution:


For solution of another solute :

Ans. Molar mass of solute = 68.23 g mol^-1.

Question 52.
What do you understand by the terms :
(1) permeable membrane
(2) semipermeable membrane?
Answer:
(1) Permeable membrane : A membrane which allows free transfer of the solute molecules from a solution of a higher concentration to a solution of a lower concentration through it is called a permeable membrane and the transfer is called diffusion, e.g., a membrane of a paper.

(2) Semipermeable membrane : A membrane which allows free passage of only the solvent molecules but not the large solute molecules or ions of large molecular mass from a solution of a lower concentration (or a pure solvent) to a solution of higher concentration through it, is called a semi-permeable membrane, e.g., parchment paper, complex like Cu₂[Fe(CN)₆], etc.

Question 53.
Define and explain osmosis.
Answer:
(1) Definition : It is defined as a spontaneous uni-directional flow of the solvent molecules from a pure solvent or a dilute solution to the more concentrated solution through a semipermeable membrane.

Example : A flow of water molecules from a dilute solution into a concentrated glucose solution through a parchment paper.

(2) Explanation : Consider a vessel divided into two compartments by a semipermeable membrane. When one compartment is filled with a pure solvent or a dilute solution and another by concentrated solution, there is a spontaneous of solvent molecules to the concentrated solution. This arises due to higher vapour pressure of a pure solvent or dilute solution than concentrated solution.

Question 54.
Explain the relation between osmotic pressure and concentration of solution.
Ans.
(1) Consider V dm³ of a solution in which n₁ moles of a solvent contains n₂ moles of a nonvolatile solute at absolute temperature T.

(2) The osmotic pressure, n of a solution is given by,
π = \(\frac{n R T}{V}\)
R is gas constant having value 0.08206 dm³ atm K^-1 mol^-1 (OR L atm K^-1 mol^-1). Since concentration, C of a solution is in mol dm^-3 or molarity is,
C = \(\frac{n}{V}\) mol dm^-3 or M
∴ π = CRT
(If concentration C is expressed in mol m^-3 and R = 8.314 J K^-1mol^-1, then π will be in SI units, pascals or Nm^-2.)

Question 55.
Explain the terms :
(1) Isotonic solutions
(2) Hypotonic solutions
(3) Hypertonic solutions.
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)
If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ARTand π_B = C_BRT where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypotonic solutions : When two solutions have different osmotic pressures, then the solution having lower osmotic pressure is said to be a hypotonic solution with respect to the other solution.
Explanation : Consider two solutions of the substances A and B having osmotic pressures π_A and π_B. If π_B is less than π_A, then the solution B is a hypotonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then, C_B < C_A. Hence, for equal volumes of the solutions, n_A < n_A.

(3) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.
Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question 56.
Explain colligative properties of electrolytes.
Answer:

• The electrolytic solutions do not exhibit colligative properties similar to nonelectrolytes.
• The colligative properties of electrolytes are higher than those shown by equimolar solutions of nonelectrolytes.
• The molar masses of electrolytes determined by colligative properties are found to be considerably lower than their actual molar masses.

Question 57.
Why are the colligative properties of electrolytic solutions greater than those for non-electrolytic solutions with same concentration ?
Answer:

• The electrolytes in polar solvents or aqueous solutions dissociate into two or more ions whereas nonelectrolytes do not dissociate.
• Consequently the number of particles in electrolytic solutions are considerably higher than equimolar nonelectrolytic solutions.
• Therefore the colligative properties of electrolytes are greater than nonelectrolytes with same concentration in solution.

Question 58.
What is abnormal colligative property? Explain the reasons.
Answer:
Abnormal colligative property : When the experimentally measured colligative property of a solution is different from that calculated theoretically by the van’t Hoff equation or by the laws of osmosis, then the solution is said to have abnormal colligative property.

Explanation : The colligative property depends on the number of solute particles in the solution but it is independent of their nature. Abnormal values of them arise when the dissolved solute undergoes a molecular change like dissociation or association in the solution.
The observed colligative property (or abnormal colligative property ) may be higher or lower than the theoretical value.

(i) Dissociation of the solute molecules : When a solute like an electrolyte is dissolved in a polar solvent like water, it undergoes dissociation, which results in the increase in the number of particles in the solution.

Hence, the observed value of the colligative property becomes higher than the theoretical value, e.g., when one mole of KCl is dissolved in the solution then due to dissociation, KCl → K⁺ + Cl^–, the number of particles increases, hence, the colligative properties like osmotic pressure elevation in the boiling point, etc. increase.

(ii) Association of the solute molecules : When a solute like a nonelectrolyte is dissolved in a nonpolar solvent like benzene, it undergoes association forming molecules of higher molecular mass. Hence, the number of the particles in the solution decreases. Therefore the colligative properties like osmotic pressure, elevation in the boiling point, etc., are lower than the theoretical value, e.g., nA → An.
2CH₃COOH → (CH₃COOH)₂
2C₆H₅COOH → (C₆H₅COOH)₂

Question 59.
Explain abnormal osmotic pressure.
Answer:

• When the experimentally observed osmotic pressure is different than theoretically calculated value by van’t Hoffs equation then it is called abnormal osmotic pressure.
• This arises when the dissolved solute undergoes a molecular change like association or dissociation.

Question 60.
Explain abnormal molecular masses.
Answer:
When the observed molecular masses obtained from their colligative properties of the substances are different (higher or lower) than the theoretical or normal values calculated from their molecular formulae, then they are called abnormal molecular masses.

Question 61.
How is van’t Hoff factor related to molecular mass of the substance ?
Answer:
The van’t Hoff factor is also defined as,
actual moles of particles in solution after


In case of nonelectrolytes, i < 1.
In case of electrolytes, i > 1. For example for KNO₃ and NaCl, i = 2, for Na₂SO₄, CaCl₂, i = 3, etc.

Question 62.
Write modification of expressions of colligative properties with the help of van’t Hoff factor.
Answer:
The modified expressions of colligative properties with the help of van’t Hoff factor i are as follows :
(1) By Raoult’s law :

Question 63.
Obtain a relation between degree of dissociation and molar mass for an electrolyte.
Answer:
Consider an electrolyte AxBy.

If i is van’t Hoff factor and n is total number of ions produced from dissociation of one mole of electrolyte then, the degree of dissociation α is given by,
α = \(\frac{i-1}{n-1}\)
Now if M_th and M_ob are theoretical arid observed molecular (or molar) masses respectively, then,
i = \(\frac{M_{\mathrm{th}}}{M_{\mathrm{ob}}}\)
∵ α(n – 1) = i – 1

This is a relation between degree of dissociation and molecular masses of the dissolved electrolyte AxBy.

Question 64.
Solve the following :

(1) 300 mL solution at 27 °C contains 0.2 mol of a nonvolatile solute. Calculate osmotic pressure of the solution.
Solution :
Given : V= 300 ml = 0.3 dm³ ;
T= 273 + 27 = 300 K
π = 0.2 mol, π = ?
π = \(\frac{n R T}{V}\)
= \(\frac{0.2 \times 0.08206 \times 300}{0.3}\)
= 16.41 atm
Ans. Osmotic pressure = π = 16.41 atm.

(2) 200 mL solution at 27 °C contains 10 g of a nonvolatile solute of molar mass 65 g mol^-1. What is osmotic pressure of the solution ?
Solution :
Given : V = 200 mL = 0.2 dm³; W = 10 g
M = 65 g mol^-1; T= 273 + 27 = 300 K; π = ?

Ans. Osmotic pressure = π = 18.93 atm.

(3) Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. (R = 8.314 J mol^-1 K^-1)
Solution :
Given : Concentration of the solution
= C = 0.2 M
= 0.2 mol dm^-3
= 0.2 × 10³ mol m^-3
Temperature = T= 300 K
= 8.314 J mol^-1 K^-1
The osmotic pressure, π is given by,
π = CRT
= 0.2 × 10³ × 8.314 × 300
= 4.988 × 10⁵ Nm^-2 (or Pa)
Ans. Osmotic Pressure = 4.988 × 10⁵ Nm^-2

(4) A solution of cane sugar containing 18 g L^-1 has an osmotic pressure 1.25 atm. Calculate the temperature of the solution. (Molar mass of cane sugar = 342, R = 0.082 lit atm mol^-1 K^-1)
Solution:
Given : Amount of cane sugar = W = 18 g L^-1
Osmotic pressure = π = 1.25 atm
Molar mass of cane sugar = M = 342 g mol^-1
Temperature = T = ?
Number of moles of cane sugar
= \(\frac{W}{M}\)
= \(\frac{18}{342}\)
= 0.05263 mol
∴ Concentration of solution = C
= \(\frac{n}{V}\)
= \(\frac{0.05263}{1}\)
= 0.05263 mol lit^-1
π = CRT
∴ T = \(\frac{\pi}{\mathcal{C R}}\)
= \(\frac{1.25}{0.05263 \times 0.08206}\)
= 289.4 K
Ans. Temperature of solution = 289.4 K

(5) Equal volumes of two solutions, one containing glucose and another urea have osmotic pressures 12.6 atm and 23.8 atm at 25 °C. Calculate the ratio of number of moles of urea to glucose.
Solution :
Given : Osmotic pressure of glucose solution = π₁
= 12.6 atm
Osmotic pressure of urea solution = π₂
= 23.8 atm
Number of moles of glucose = n₁
Number of moles of urea = n₂

(6) One litre of a solution containing 25 g glucose has osmotic pressure 3.4 atm at 300 K. If by diluting the solution the osmotic pressure becomes 2.0 atm at the same temperature, what would be its concentration ? (Molar mass of glucose = 180)
Solution :
Given : Mass of solute (glucose) = 25 g
Volume of solution = 1.0 L
Osmotic pressure = π₁ = 3.4 atm
After dilution osmotic pressure = π₂ = 2 atm
Final concentration = c₂ = ?
The number of moles of glucose = n = \(\frac{25}{180}\)
= 0.1389 mol
Initial concentration = c₁ = \(\frac{n}{V}\)
= \(\frac{0.1389}{1}\)
At constant temperature, by van’t Hoff-Boyle’s law,

Ans. Concentration of solution = 0.0817 mol L^-1

(7) A solution of a substance having mass 1.8 × 10^-3 kg has the osmotic pressure of 0.52 atm at 280 K. Calculate the molar mass of the substance used.
[Volume = 1 dm³, R = 8.314 JK^-1 mol^-1]
Solution :
Given : W= 1.8 × 10^-3 kg = 1.8 g
π = 0.52 atm; V = 1 dm³
T = 280 K;
Molar mass = M = ?

Ans. Molar mass = M = 79.53 g mol^-1

(8) An organic substance (M = 169 gram mol^-1) is dissolved in 2000 cm³ of water. Its osmotic pressure at 12 °C was found to be 0.54 atm. If R = 0.0821 L atm K^-1 mol^-1, calculate the mass of the solute.
Solution :
Given : M = 169 g mol^-1
V = 200 cm3 = 0.2 dm³
T = 273 + 125 = 285 K
π = 0.54 atm
R = 0.0821 L atm K^-1 mol^-1; W = ?

Ans. Mass of solute = 0.78 g

(9) A solution containing 10 g glucose has osmotic pressure 3.84 atm. If 10 g more glucose is added to the same solution, what will be its osmotic pressure. (Temperature remains constant)
Solution :
Given : Mass of glucose = W₁ = 10 g
Mass of more glucose added = W₂ = 10 g
Initial osmotic pressure = π₁ = 3.84 atm
Final osmotic pressure = π₂ = ?
Total mass of glucose in final solution = W₁ + W₂
W’ = 10 + 10
= 20 g

∴ π₂ = \(\pi_{1} \times \frac{W^{\prime}}{W_{1}}\)
= \(3.84 \times \frac{20}{10}\) = 7.68 atm
Ans. Osmotic pressure of the solution = 7.68 atm

(10) A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Solution :
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}=2.312\)
Ans. van’t Hoff factor = i = 2.312

Multiple Choice Questions

Question 65.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. A molal solution is one that contains one mole of solute in
(a) one litre of the solvent
(b) 1000 g of the solvent
(c) one litre of the solution
(d) 22.4 litre of solution
Answer:
(b) 1000 g of the solvent

2. 10.0 grams of caustic soda when dissolved in 250 cm³ of water, the resultant molarity of solution is
(a) 0.25 M
(b) 0.5 M
(c) 1.0 M
(d) 0.1 M
Answer:
(c) 1.0 M

3. 5 × 10-3 kg of urea is dissolved in 2 × 10^-2 kg of water. The percentage by weight of urea is
(a) 15%
(b) 20%
(c) 25%
(d) 30%
Answer:
(b) 20%

4. Vapour pressure of solution of a nonvolatile solute is always
(a) equal to the vapour pressure of pure solvent
(b) higher than vapour pressure of pure solvent
(c) lower than vapour pressure of pure solvent
(d) constant
Answer:
(c) lower than vapour pressure of pure solvent

5. According to the Raoult’s law, the relative lowering of vapour pressure is equal to the
(a) mole fraction of solvent
(b) mole fraction of solute
(c) independent of mole fraction of solute
(d) molality of solution
Answer:
(b) mole fraction of solute

6. Partial pressure of solvent in solution of non-volatile solute is given by equation,
(a) P = x₂P⁰
(b) P⁰ = xP
(c) P = x₁P⁰
(d) P⁰ = x₁P
Answer:
(c) P = x₁P⁰

7. When partial pressure of solvent in solution of nonvolatile solute is plotted against its mole fraction, nature of graph is
(a) a straight line passing through origin
(b) a straight line parallel to mole fraction of solvent
(c) a straight line parallel to vapour pressure of solvent
(d) a straight line intersecting vapour pressure axis
Answer:
(a) a straight line passing through origin

8. Colligative property depends only on ………………. in a solution.
(a) Number of solute particles
(b) Number of solvent particles
(c) Nature of solute particles
(d) Nature of solvent particles
Answer:
(a) Number of solute particles

9. Lowering of vapour pressure of solution
(a) is a property of solute
(b) is a property of solute as well as solvent
(c) is a property of solvent
(d) is a colligative property
Answer:
(d) is a colligative property

10. Raoult’s law is not applicable to
(a) solutions of volatile solutes
(b) solutions of electrolytes
(c) dilute solutions
(d) concentrated solutions
Answer:
(d) concentrated solutions

11. The relative lowering of vapour pressure of a solution is proportional to the
(a) mole fraction of the solvent
(b) mole fraction of the solute
(c) amount of the substance
(d) volume of the solvent
Answer:
(b) mole fraction of the solute

12. The vapour pressure of an aqueous solution of glucose at 100 °C is 710 mm Hg. Hence the molality of the solution is
(a) 2.83 m
(b) 3.65 m
(c) 16.47 m
(d) 12.5 m
Answer:
(b) 3.65 m

13. Relative vapour pressure lowering depends only on
(a) Mole fraction of solute
(b) Nature of solvent
(c) Nature of solute
(d) Nature of solute and solvent
Answer:
(a) Mole fraction of solute

14. A solution having mole fraction of a solute equal to 0.05 has vapour pressure 20 × 10³ Nm^-2. Hence the vapour pressure of a pure solvent is
(a) 21.05 × 10³ Nm^-2
(b) 4 × 10⁵ Nm^-2
(c) 1 × 10³ Nm^-2
(d) 2 × 10⁴ Nm^-2
Answer:
(a) 21.05 × 10³ Nm^-2

15. The addition of the nonvolatile solute into the pure solvent ……………..
(a) increases the vapour pressure of solvent
(b) decreases the boiling point of solvent
(c) decreases the freezing point of solvent
(d) increases the freezing point of solvent
Answer:
(c) decreases the freezing point of solvent

16. A solution having the highest vapour pressure is
(a) 1 M Al₂(SO₄)₃
(b) 0.1 M NaNO₃
(c) 1 M BaCl₂
(d) 1 M Ca(NO₃)₂
Answer:
(b) 0.1 M NaNO₃

17. Molal elevation constant is elevation in boiling point produced by
(a) 1 g of solute in 100 g of solvent
(b) 100 g of solute in 1000 g of solvent
(c) 1 mole of solute in one litre of solvent
(d) 1 mole of solute in one kg of solvent
Answer:
(d) 1 mole of solute in one kg of solvent

18. The determination of molar mass from elevation in boiling point is called
(a) cryoscopy
(b) osmometry
(c) ebullioscopy
(d) spectroscopy
Answer:
(c) ebullioscopy

19. Which of the following aqueous solutions will have minimum elevation in boiling point ?
(a) 0.1 M KCl
(b) 0.05 M NaCl
(c) 1 M AIPO₄
(d) 0.1 M MgSO₄
Answer:
(b) 0.05 M NaCl

20. Which of the following solutions shows maximum depression in freezing point ?
(a) 0.5 M Li₂SO₄
(b) 1 M NaCl
(c) 0.5 M Al₂ (SO₄)₃
(d) 0.5 M BaCl₂
Answer:
(c) 0.5 M Al₂ (SO₄)₃

21. If the freezing point of 0.1 m solution is 272.814 K, then the freezing point of 0.2 m solution will be
(a) 545.628 K
(b) 265.64 K
(c) 272.628 K
(d) 0.482 K
Answer:
(c) 272.628 K

22. At a freezing point,
(a) Vapour pressure of a solution = Vapour pressure of a solid
(b) Vapour pressure of a liquid = Vapour pressure of a solid
(c) Vapour pressure of a liquid > Vapour pressure of a solid
(d) Vapour pressure of a solid > Vapour pressure of a liquid
Answer:
(b) Vapour pressure of a liquid = Vapour pressure of a solid

23. In osmosis
(a) solvent molecules pass from high concentration of solute to low concentration
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute
(c) solute molecules pass from low concentration to high concentration
(d) solute molecules pass from high concentration to low concentration
Answer:
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute

24. As temperature increases
(a) Osmotic pressure and vapour pressure decrease
(b) Vapour pressure and osmotic pressure increase
(c) Vapour pressure increases but osmotic pressure decreases
(d) Osmotic pressure increases but vapour pressure decreases
Answer:
(b) Vapour pressure and osmotic pressure increase

25. A mango kept in a salt solution shrinks. Hence the liquid content in mango with respect to the salt solution is
(a) isotonic
(b) hypertonic
(c) hypotonic
(d) equimolar
Answer:
(c) hypotonic

26. The osmotic pressure of 5% glucose solution at 300 K is
(a) 6.93 × 10⁵ Nm^-2
(b) 6.93 × 10² Nm^-2
(c) 6.83 Pa
(d) 6.00 × 10³ Pa
Answer:
(a) 6.93 × 10⁵ Nm^-2

27. 0.1 M solution of A has osmotic pressure xNm^-2 at 300 K. If 200 ml of A and 100 ml of 0.2 M solution of nonreactive solute B are mixed then the osmotic pressure will be
(a) 3x
(b) 0.05 x
(c) 1.33 x
(d) 0.75 x
Answer:
(c) 1.33 x

28. The solution A is twice hypertonic to the solution B at a given temperature. If the solution A contains 8.6 × 10²² molecules, then the number of molecules present in B are,
(a) 8.6 × 10²²
(b) 1.73 × 10²³
(c) 3.24 × 10²²
(d) 4.3 × 10²²
Answer:
(d) 4.3 × 10²²

29. Isotonic solutions are the solutions having the same
(a) surface tension
(b) vapour pressure
(c) osmotic pressure
(d) viscosity
Answer:
(c) osmotic pressure

30. If Kb for water is 0.52 Km^-1, the boiling point of 0.2 m solution of a nonvolatile solute will be
(a) 371.96 K
(b) 373.104 K
(c) 373.52 K
(d) 374.0 K
Answer:
(b) 373.104 K

31. The vapour pressure of water is 15.5 mm at 20 °C. The lowering of vapour pressure of 0.02 m K Br solution will be
(a) 0.0112 mm Hg
(b) 0.0056 mm Hg
(c) 0.056 mm Hg
(d) 0.31 mm Hg
Answer:
(a) 0.0112 mm Hg

32. Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure ?
(a) Al₂(SO₄)₃
(b) Na₂SO₄
(c) MgCl₂
(d) KCl
Answer:
(a) Al₂(SO₄)₃

33. If equimolar solutions of urea, NaCl, sucrose and BaCl₂ have boiling points A, B, C and D, then
(a) A = C < B < D
(b) A = D < B < C
(c) A > B > C < D
(d) A < B < C < D
Answer:
(a) A = C < B < D

34. The molar mass of acetic acid obtained by measuring depression in freezing point is 115.8 gmol^-1. Hence the degree of association is
(a) 0.482
(b) 0.964
(c) 0.883
(d) 1.12
Answer:
(b) 0.964

35. Δ T_b/m for NaBr solution will have value (K_b = 0.52 K kg mol^-1)
(a) 0.52 K mol^-1
(b) 0.104 K kg mol^-1
(c) 1.24 kg mol^-1
(d) 1.04 K kg mol^-1
Answer:
(d) 1.04 K kg mol^-1

36. 0.2 M urea solution can be isotonic with
(a) 0.1 M KBr solution
(b) 0.1 M glucose solution
(c) 0.15 m NaCl solution
(d) 0.2 M KBr solution
Answer:
(a) 0.1 M KBr solution

37. The osmotic pressure of 0.1 M HCl solution at 27 °C will be
(a) 2.46 atm
(b) 0.164 atm
(c) 4.92 atm
(d) 0.0082 atm
Answer:
(c) 4.92 atm

38. If a, b, c and d are the van’t Hoff factors for Na₂SO₄, glucose and K₄[Fe(CN)₆] then
(a) a > b > c
(b) a < b < c
(c) b < a < c
(d) c < a < b
Answer:
(c) b < a < c

39. The osmotic pressure of 0.2 M KCl solution at 310 K is
(a) 10.17 atm
(b) 5.084 atm
(c) 8.36 atm
(d) 12.2 atm
Answer:
(a) 10.17 atm

40. A temperature at which 0.1 M KCl solution will have osmotic pressure 10 atm will be
(a) 408 °C
(b) 263 °C
(c) 310 °C
(d) 337 °C
Answer:
(d) 337 °C

41. 5 % solution of glucose is isotonic with a solution of urea (M = 60). Hence the weight of urea present in the solution is
(a) 1.67 g
(b) 6.0 g
(c) 18.6 g
(d) 1.2 g
Answer:
(a) 1.67 g

42. Abnormal molar mass is produced by
(a) association of solute
(b) dissociation of solute
(c) both association and dissociation of solute
(d) separation by semipermeable membrane
Answer:
(c) both association and dissociation of solute

43. The van’t Hoff factor for an aqueous solution of an electrolyte is
(a) less than 1
(b) zero
(c) greater than 1
(d) equal to 1
Answer:
(c) greater than 1

44. The value of van’t Hoff factor will be minimum for
(a) 0.05 M AlCl₃
(b) 0.2 M NaNO₃
(c) 5.0 M glucose
(d) 0.1 M H₂SO₄
Answer:
(c) 5.0 M glucose

45. van’t Haff factor for K₄[FeC(N)₆] dissociated 10% is
(a) 1.1
(b) 1.4
(c) 0.86
(d) 1.6
Answer:
(b) 1.4

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 8 Company Meetings – II Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 8 Company Meetings – II

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Annual General Meeting is held _________ in a year.
(a) once
(b) twice
(c) thrice
Answer:
(a) once

Question 2.
Annual General Meeting is convened by _________
(a) Creditors
(b) Board of Directors
(c) Committee
Answer:
(b) Board of Directors

Question 3.
First Annual General Meeting shall be held within _________ months of the closing of the financial year.
(a) 6
(b) 9
(c) 15
Answer:
(b) 9

Question 4.
The quorum for the Annual General Meeting of a public company with 10,000 shareholders is _________ members.
(a) 5
(b) 30
(c) 15
Answer:
(b) 30

Question 5.
First Board Meeting is held within _________ days of the date of its incorporation.
(a) 30
(b) 15
(c) 45
Answer:
(a) 30

Question 6.
The Gap between two Board Meetings should not be more than _________ days.
(a) 30
(b) 60
(c) 120
Answer:
(c) 120

Question 7.
With the help of technology _________ meeting is held for Board of Directors.
(a) Visual
(b) Virtual
(c) Audio
Answer:
(b) Virtual

Question 8.
The Secretary has to get the signatures of the members present at the meeting in _________ ‘Attendance Book’.
(a) Directors
(b) Members
(c) Creditors
Answer:
(b) Members

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Directors	(1) 21 days
(b) Annual General Meeting	(2) Statutory Report
(c) Auditor	(3) Preference shareholders
(d) Notice – Board Meeting	(4) Extra-Ordinary General Meeting
(e) Class Meeting	(5) General Meeting
	(6) 7 days
	(7) Annual Report
	(8) Debenture holders
	(9) Annual General Meeting
	(10) Committee Meeting

Answer:

Group ‘A’	Group ‘B’
(a) Directors	(10) Committee Meeting
(b) Annual General Meeting	(7) Annual Report
(c) Auditor	(9) Annual General Meeting
(d) Notice – Board Meeting	(6) 7 days
(e) Class Meeting	(3) Preference shareholders

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A meeting in which Directors are elected.
Answer:
Annual General Meeting

Question 2.
A meeting of Shareholders is held after the financial year is over.
Answer:
Annual General Meeting

Question 3.
A meeting of the Board of Directors was held with the help of technology.
Answer:
Virtual Meeting

Question 4.
The quorum for Annual General Meeting of a public company with 4000 shareholders.
Answer:
15 Members

Question 5.
The Gap between two Annual General Meetings should not be more than.
Answer:
15 months

Question 6.
The quorum for Board Meetings.
Answer:
1/3rd rd or 2 directors (whichever is higher)

Question 7.
The Gap between the two Board Meetings should not be more than.
Answer:
120 days

1D. State whether the following statements are True or False.

Question 1.
30 clear days’ notice should be given in case of the Annual General meeting.
Answer:
False

Question 2.
Shareholders can appoint a proxy to attend and vote at the meeting on their behalf.
Answer:
True

Question 3.
Board Meetings are called once a year.
Answer:
False

Question 4.
Debenture holders are the owners of the company.
Answer:
False

Question 5.
Directors have to perform duties related to Annual General Meeting.
Answer:
False

Question 6.
Quorum is just before the commencement of the meeting.
Answer:
True

Question 7.
Secretary takes down the notes of the proceedings of the meeting.
Answer:
True

1E. Find the odd one.

Question 1.
Shareholders Meeting, Directors Meeting, Committee Meeting.
Answer:
Shareholders meeting

Question 2.
Board Meeting, Committee Meeting, Class Meeting.
Answer:
Class Meeting

Question 3.
AGM, EOGM, Committee Meeting.
Answer:
Committee Meeting

Question 4.
General Meeting, Class Meeting, Creditors Meeting.
Answer:
Creditors Meeting

Question 5.
Debenture holders Meeting, Creditors Meeting, Class Meeting.
Answer:
Class Meeting

1F. Complete the sentences.

Question 1.
Auditor is appointed in _________
Answer:
Annual General Meeting

Question 2.
An Extra-Ordinary General Meeting is held _________
Answer:
under special circumstances

Question 3.
Minutes of the meeting is the duty of _________
Answer:
Secretary

Question 4.
Directors are appointed in _________
Answer:
Annual General Meeting

Question 5.
Minutes of the meeting has to be drafted by the secretary in _________
Answer:
15 days

Question 6.
Auditor is appointed for a period of _________
Answer:
1 year

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) ……………………..	Annual General Meeting
(2) Proxy	……………………….
(3) …………………….	Annual Report
(4) Minutes	……………………….
(5) ……………………	Notes of the proceedings of a meeting
(6) Notice of Board Meeting	……………………….

(Shareholder, 15 days, Secretary, 7 days, Auditor, Annual General Meeting)
Answer:

Group ‘A’	Group ‘B’
(1) Auditor	Annual General Meeting
(2) Proxy	Shareholder
(3) Annual General Meeting	Annual Report
(4) Minutes	15 days
(5) Secretary	Notes of the proceedings of a meeting
(6) Notice of Board Meeting	7 days

1H. Correct the underlined word and rewrite the following sentences.

Question 1.
In a Partnership Firm, there is the separation of ownership and management.
Answer:
In a Joint Stock Company, there is the separation of ownership and management.

Question 2.
The interval between 2 Annual General Meetings should be no more than 12 months.
Answer:
The interval between 2 Annual General Meetings should be no more than 15 months.

Question 3.
Annual General Meeting is called to transact “special business.”
Answer:
Extra Ordinary General Meeting is called to transact “special business.”

Question 4.
Meetings of debenture holders, depositors, etc. are called Shareholders Meeting.
Answer:
Meetings of debenture holders, depositors, etc. are called Creditor’s Meeting.

2. Explain the following terms/concepts.

Question 1.
Class Meeting
Answer:
A Meeting of a particular class of shareholders, held to make changes in their rights and duties, is called a class meeting. It is held to get their consent for the changes affecting their interest.

Question 2.
Annual General Meeting (AGM)
Answer:
Meeting of equity shareholders which are held once every year is called AGM. It is held to review and discuss the progress made by the company during the financial year. It is held to take decisions like:

• to adopt Annual a/c, Director’s Report, and Auditor’s Report
• to declare dividend
• to elect directors
• to appoint auditors and fix their remuneration.

Question 3.
Extra Ordinary General Meeting
Answer:
A meeting held to discuss and decide special or urgent matters like alteration in MOA or AOA, removal of directors or auditor before expiry of his term, voluntary winding up of company, etc is called an extraordinary general meeting. Such meetings can be called by the Board of Directors or by BOD on the requisition of members or by Requisitionist themselves or by National Company Law Tribunal (NCLT) or Government.

Question 4.
Director’s Meeting
Answer:
The meeting which is held to discuss the policy matters of the company is called as Director’s meeting. Such meetings are classified into Board meetings and Committee meetings. The First Board meeting is to be called within 30 days from the date of incorporation.

3. Answer in brief.

Question 1.
State any four legal provisions relating to Extra-Ordinary General Meeting.
Answer:
Legal Provisions relating to Extra-Ordinary General Meeting:
(i) Time for holding a meeting:
Extra-Ordinary General Meeting is between two Annual General Meetings under special circumstances. It can be held at any time as per the requirements of the company.

(ii) Authority to Convene:

• The Board of Directors has the right to call an Extra-Ordinary General Meeting by sending a proper notice to the shareholders.
• Extra-Ordinary General Meeting can be called by the members holding at least 1/10th of the paid-up capital or 1/10th of voting power in the company. Board must call Extra-Ordinary General Meeting within 45 days of receiving the requisition from the members.
• If the Board fails to call such a meeting then the requisitionists themselves call this meeting within 3 months from the date of deposit of the requisition. The company shall pay all the expenses incurred for holding Extra-ordinary General Meeting.
• National Company Law Tribunal (NCLT) can order such meetings on its own or at the request of a director or any members having voting rights.

(iii) Notice:
The notice must be given to all those who are entitled to receive it, at least 21 clear days in advance of the meeting. It is sent to the members at their registered address by post or through electronic mode like Email, etc.

(iv) Quorum:
According to the Companies Act, 2013, the quorum for Extra-Ordinary General Meeting of a public company is as follows:

No. of Shareholders	Quorum
Upto 1000	5 Members
1000 – 5000	15 Members
More than 5000	30 Members

For a private company minimum of two members should be present in person.

4. Answer the following questions.

Question 1.
Explain the functions of a secretary related to the Extra-Ordinary General Meeting.
Answer:
Functions (duties) of a secretary related to Extra Ordinary General Meeting:
Duties Before the Meeting:

• Board Meeting: The secretary has to convene a board meeting in consultation with the Chairman. The board has to decide the day, date, time, and place of the meeting.
• Notice: The secretary drafts the notice, agenda, and other required documents. He gets it approved by the chairman. Then he sends 21 clear days’ notice and other documents to the members before the meeting.
• Public Notice: He has to publish notice of the extraordinary general meetings in leading newspapers for the information of the general public.
• Arrangement: The secretary has to make arrangements such as booking the hall, refreshments, documents, etc.

Duties During the Meeting:

• Quorum: The secretary checks the quorum at the beginning of the meeting and informs the same of the chairman.
• Attendance: The secretary has to get signatures of members present at the meeting in ‘Members Attendance Book’.
• Reading Notice: The secretary reads out the notice and agenda of the meeting as per the direction of the chairman.
• Assist Chairman: The secretary assists the chairman by providing necessary information, records, document, etc.
• Notes of Proceedings: The secretary should take down the notes of the proceedings of the meeting. He has to draft the minutes on the basis of these notes.
• Poll: The secretary has to make necessary arrangements for taking poll if it is demanded.

Duties After the Meeting:

• Drafting Minutes: The secretary drafts the minutes of the meeting on the basis of the notes taken down during the proceedings of the meeting.
• Implementation of Decisions: The secretary instructs the concerned department for implementing the decisions taken at the meeting.
• Filing Resolutions: The secretary has to file the certified copy of resolutions passed at the meeting to the Registrar of Companies.

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 2 Systematics of Living Organisms Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 2 Systematics of Living Organisms

Question 1.
Write the definition of systematics given by G. Simpson in 1961.
Answer:
Systematics is the study of kinds and diversity of organisms and their comparative and evolutionary relationship.

Quesiton 2.
Explain the term taxonomy.
Answer:

1. Taxonomy means classification following certain rules or principles.
2. The word taxonomy comes from two Greek words, taxis meaning arrangement and nomous meaning law or rule.
3. The term taxonomy was coined by A.P. de Candolle (Swiss Botanist) [1778-1841].

Question 3.
Who coined the term taxonomy?
Answer:
The term taxonomy was coined by A.P. de Candolle (Swiss Botanist) [1778-1841].

Question 4.
Define the term classification. What is the basis of classification?
Answer:
1. Classification is the arrangement of organisms or groups of organisms in distinct categories in accordance with a particular and well-established plan.
2. It is based on the similarities and differences among the organisms.

Question 5.
What are the three types of classification systems?
Answer:
The three types of classification systems are:
(i) Artificial system:
(a) It is based on few visible, easily observable characters, which are non-evolutionary such as habit, colour, form, etc.
(b) It does not consider the affinities (relationships) among different organisms.
E.g. Linnaeus system of classification.

(ii) Natural system:
It is based on objectively significant characters with respect to their affinities with other organisms.
E.g. Bentham and Hooker’s system of classification.

(iii) Phylogenetic system:
It is based on the phylogenetic relationship between different organisms with respect to common evolutionary descent (ancestor).
E.g. Engler and Prantl’s classification.

Question 6.
What is domain? Name the three domains of life.
Answer:
1. Domain is a unit larger than Kingdom in the system of classification.
2. Three domains of life are Archaea, Bacteria and Eukarya.

Question 7.
Who proposed the three-domain system?
Answer:
Carl Woese proposed the three-domain system.

Question 8.
State one similarity and difference between archaea and bacteria?
Answer:
Both archaea and bacteria are prokaryotic. They differ in their cell wall structures.

Question 9.
Which domain has eukaryotic cells?
Answer:
Domain Eukarya has eukaryotic cells.

Question 10.
what is chemotaxonomy? Explain with example.
Answer:

1. It is method of biological classification based on the similarities and differences in structure of certain chemical compounds present among the organisms being classified.
2. Thus, it is a classification based on chemical constituents of organisms.
3. For e.g. Cell wall with peptidoglycan is present in Bacteria while it is absent in Archaea. Among Eukarya, fungi have chitinous cell wall, while plants have cellulosic cell wall.

Question 11.
Write a short note on numerical taxonomy.
Answer:
Numerical taxonomy:

1. It is based on quantification of characters and develops an algorithm for classification.
2. The aim of this was to create a taxonomy using numeric algorithms like cluster analysis rather than using subjective evaluation of their properties.
3. It was proposed by Sokel and Sneath in 1963.

Question 12.
What is cladogram? Give a diagrammatic representation of three domains of life with the help of cladogram.
Answer:
1. It is a representation of hypothetical relationship denoting a comparison of organisms and their common ancestors.
2. It has a typical branching pattern.

Question 13.
Write in detail about the Phylogeny.
Answer:
Phylogeny:

1. It is the evolutionary relationship of organism.
2. It is an important tool in classification as it considers not merely the morphological status but also the relationship of one group of organisms with other groups of life.
3. The system helps to understand the evolution and also focuses on the similarities of their metabolic functioning.
4. Woese’s three domain concept as well as Whittaker’s five kingdom system are examples of phylogenetic relationship.

Question 14.
What is the use of DNA barcoding?
Answer:
DNA barcoding helps to study newly identified species as well as understanding ecological and evolutionary relationships between living organisms.

Question 15.
What are the steps involved in the process of DNA barcoding?
Answer:
The process of DNA barcoding includes two basic steps:
1. Collecting DNA barcode data of known species.
2. Matching the barcode sequence of the unknown sample against the barcode library for identification.

Question 16.
What are the applications of DNA barcoding?
Answer:
The applications of DNA barcoding are as follows:

1. It helps to protect endangered species.
2. It plays an important role in preservation of natural resources.
3. It is also used for pest control in agriculture.
4. It is used for identification of disease vectors.
5. It is used for authentication of natural health products.
6. It is also used for identification of medicinal plants.

Question 17.
What is taxonomic category?
Answer:

1. Category is a rank or level in the hierarchial classification of organisms.
2. Each category is referred to as a unit of classification.
3. Category is a part of taxonomic arrangements hence, called taxonomic category.
4. All categories together constitute the taxonomic hierarchy.

Question 18.
What are the compulsory taxonomic categories?
Answer:
Kingdom, division, class, order, family, genus, species are the compulsory categories.

Question 19.
What are the facultative taxonomic categories?
Answer:
Sub-order, sub-family, etc. are the facultative categories which are used when required.

Question 20.
Define taxonomic hierarchy.
Answer:
The manner of scientific grouping of different taxonomic categories in a descending order on the basis of their ranks or positions in classification is called taxonomic hierarchy.

Question 21.
Define the term Taxon. Give some examples of taxa at different hierarchical levels.
Answer:
1. Taxon is a group of living organisms of any rank in the system of classification.
2. In plant kingdom, each taxonomic group such as angiospermae, dicotyledonae, polypetalae, malvaceae represents a taxon.

Question 22.
Write the classification of:
Answer:
1. China Rose
2. Cobra

Question 23.
Explain the following terms:

1. Species
2. Genus
3. Family
4. Order
5. Class
6. Division/Phylum
7. Sub kingdom
8. Kingdom

Answer:
(i) Species:
(a) Species is the principal natural taxonomic unit, ranking below a genus.
(b) It is a group of organisms that can interbreed under natural condition to produce fertile offspring.
(c) It was thought to be an indivisible, stable and static unit.
(d) However, in the modem taxonomy, subdivision of species such as sub-species, varities and populations are seen and given more importance.

(ii) Genus:

(a) Genus is a taxonomic rank or category larger than species used in the biological classification of living and fossil organisms.
(b) Genus is a group of species bearing close resemblance to one another in their morphological characters but they do not interbreed.
(c) For e.g. Tiger, Leopard, Lion all three belong to same genus Panthera. They have common characters yet are different from each other because their genus is same but species is different.
(d) Another example is genus Solarium. Brinjal and potato both belong to this genus.

(iii) Family:

(a) It is one of the major hierarchial taxonomic rank.
(b) A family represents a group of closely related genera.
(c) For e.g. genera like Hibiscus, Gossypium, Sida, Bombax are included in same family Malvaceae.
(d) Although, there are many similarities between cat and dog, cat belongs to the family of leopards, tigers and lions, i.e. family Felidae and dog belongs to different family i.e. Canidae.

(iv) Cohort/Order:

(a) It is taxonomic rank used in the classification of organisms and recognised by nomenclature codes.
(b) An order is a group of closely related families showing definite affinities.
(c) Members belonging to same order but different families may show very few dissimilarities.
(d) For e.g. family Papaveraceae, Brassicaceae, Capparidaceae, etc with parietal placentation are grouped in order Parietales.
(e) Families of dogs and cats though are different, they belong to same order Carnivora.

(v) Class:

(a) The class is the distinct taxonomic rank of biological classification having its own distinctive name.
(b) Class is the assemblage of closely allied orders.
(c) For e.g. Orders Carnivora and order Primates belong to class Mammalia. Thus monkeys, gorillas, gibbons (Primates) and dogs, cats, tigers (Carnivora) belong to same class.

(vi) Division/ Phylum:

(a) The division is a category composed of related classes.
(b) For e.g. division Angiospermae includes two classes Dicotyledonae and Monocotyledonae.
(c) In animal classification, instead of division, the category Phylum is used.

(vii) Sub-kingdom:

(a) Different divisions having some similarities form sub-kingdom.
(b) The divisions Angiospermae and Gymnospermae forms the sub-kingdom Phanerogams or Spermatophyta (all seed producing plants).

(viii) Kingdom:

(a) It is the highest taxonomic category composed of different sub-kingdoms.
(b) For e.g. sub-kingdom Phanerogams and Cryptogams form the Plant kingdom or Plantae which includes all the plants, while all animals are included in kingdom Animalia.

Question 24.
Define nomenclature.
Answer:
The art of giving name to the organism is called nomenclature.

Question 25.
What is meant by vernacular name?
Answer:
Vernacular names are the names which are given to organisms in a particular region and language by local people.

Question 26.
What are the disadvantages of vernacular names of organisms?
Answer:
Disadvantages of vernacular names/ local names/ common names:

1. Vernacular names do not indicate the necessary information about the organism.
2. It does not indicate proper relationship of the organisms.
3. Vernacular names are not universal, e.g. Pansy (Viola tricolor L.) grown in most European and American gardens has about 50 common english names. In Ayurveda, mango (Mangifera indica L.) is known by over 50 different names which are in Sanskrit language.
4. Vernacular names have limited usage.
5. Local names are different and confusing.

Question 27.
Who proposed binomial system of nomenclature?
Answer:
Swedish naturalist Carl Linnaeus proposed binomial system of nomenclature.

Question 28.
What is binomial nomenclature? Give the rules for binomial nomenclature.
Answer:
1. A system of nomenclature of plants and animals in which the scientific name consists of two words or parts or epithets is called binomial nomenclature.
2. This system of nomenclature was developed by Carl Linnaeus. He gave certain principles for this nomenclature in his book ‘Species Plantarum’.

Rules of binomial nomenclature:

1. The name of the organism is composed of two Latin or Greek words.
2. Generic epithet is a simple noun which should come first and always begin with a capital letter.
3. Specific epithet is the descriptive adjective which should come later and begin with a small letter.
4. The generic and specific epithet must be underlined separately if hand written or in italics when printed.
5. The generic as well as specific epithet should not have less than three letters and more than thirteen letters.
6. Usually the name of the author who names a plant or animal is also written in full or abbreviated form after scientific name. e.g. Mangifera indica L. Where L stands for Linnaeus.

Question 29.
In Mangifera indica L., what does letter ‘L’ indicate?
Answer:
In Mangifera indica L., letter L indicates author’s name i.e. Linnaeus.

Question 30.
Which kingdoms were included in two kingdom system of classification? Who introduced it?
Answer:
The two-kingdom system of classification included Kingdom plantae and Kingdom animalia. This system was introduced by Carl Linnaeus.

Question 31.
What was the drawback of two kingdom system of classification?
Answer:
Two kingdom system was found inadequate for classification of some organisms like bacteria, fungi, Euglena, etc.

Question 32.
Who suggested five kingdom system of classification?
Answer:
R.H. Whittaker suggested five kingdom system of classification.

Question 33.
Match the following.

Column I	Column II
1. Vibrio	a. Rod-shaped
2. Bacillus	b. Spherical
3. Spirillum	c. Spiral shaped
	d. Comma or kidney-shaped

Answer:

Column I	Column II
1. Vibrio	d. Comma or kidney-shaped
2. Bacillus	a. Rod-shaped
3. Spirillum	c. Spiral shaped

Question 34.
Identify the different shapes of bacterial cells shown in the given figures:

Answer:
Figure a: Coccus;
Figure b: Coccobacillus;
Figure c: Vibrio
Figure d: Bacillus;
Figure e: Spirillum; Figure f: Spirochete

Question 35.
What are Archaebacteria?
Answer:

1. These are the most primitive type of bacteria.
2. They are differentiated from other bacteria on the basis of their different cellular features.
3. These bacteria are mostly found in the extreme environmental conditions, hence called extremophiles.
4. Bacteria that can withstand high salinities are called halophiles, while those that withstand extreme temperature are known as thermophiles.
5. Methanogenic bacteria found in gut of ruminants (cows and buffaloes) help in production of methane in biogas plants.

Question 36.
Why are archaebacteria called extremophiles?
Answer:

• These bacteria are mostly found in the extreme environments, hence called extremophiles.
• They have capacity to survive in very severe conditions.
• They are found in a variety of places from volcanic craters to salty lakes and hot springs.

Question 37.
Write in detail about Eubacteria.
Answer:
Eubaceria:

1. These are commonly referred as true bacteria.
2. They have cell wall made up of peptidoglycan.
3. Eubacteria are mostly heterotrophic, few are autotrophic.
4. The autotrophs can be photosynthetic like Chlorobium (Green sulphur bacteria) and Chromatium or chemosynthetic like sulphur bacteria.
5. These are mostly multicellular filamentous forms living in fresh water.
6. Filaments show heterocyst which helps in nitrogen fixation.
7. The body is covered by mucilaginous sheath.
8. The genetic material is typical prokaryotic.
9. The photosynthetic pigments include Chl-a, Chl-b, carotenes and xanthophylls.
10. Most of them are decomposers that help in breaking down large molecules in simple molecules or minerals.

Question 38.
Write a short note on useful and harmful bacteria.
Answer:
(i) Useful bacteria:
Most of the bacteria act as a decomposer. They breakdown large molecules in simple molecules or minerals. Examples of some useful bacteria:
Lactobacillus’. It helps in curdling of milk.
Azotobacter. It helps to fix nitrogen for plants.
Streptomyces: It is used in antibiotic production such as streptomycin.
Methanogens: These are used for production of methane (biogas) gas from dung.
Pseudomonas spp. and Alcanovorax borkumensis: These bacteria have the ability to destroy the pyridines and other chemicals. Hence, used to clear the oil spills.

(ii)Harmful bacteria:
This includes disease causing bacteria. They cause various diseases like typhoid, cholera, tuberculosis, tetanus, etc. Examples of some harmful bacteria:
Salmonella typhi: It is a causative organism of typhoid.
Vibrio cholerae: It causes cholera.
Mycobacterium tuberculosis’. It causes tuberculosis.
Clostridium tetani: It causes tetanus.
Clostridium spp.: It causes food poisoning.
Many forms of mycoplasma are pathogenic.
Agrobacterium , Erwinia, etc are the pathogenic bacteria causing plant diseases.
Animals and pets also suffer from bacterial infections caused by Brucella, Pastrurella, etc.

Question 39.
Identify label X and Y in the given figure of Cyanobacteria (Nostoc).
Answer:

Question 40.
what is Mycoplasma?
Answer:

1. These are the smallest living cells known.
2. They lack cell wall.
3. Many forms are pathogenic.
4. They are resistant to common antibiotics because they lack cell wall.

Question 41.
Identify the following diagram, label it and write detail information in your words.
Answer:
The given figure represents Paramoecium.
Characteristics:

1. It belongs to kingdom Protista. It is further classified as animal like protist.
2. It lacks cell wall.
3. It shows heterotrophic and holozoic nutrition.
4. It is a ciliated protozoan where locomotion is due to cilia.
5. It has gullet (a cavity) which opens on the cell surface.

Question 42.
Which kingdom shows link with all eukaryotic members?
Answer:
Kingdom Protista shows link with all eukaryotic kingdoms such as kingdom plantae, fungi and animalia.

Question 43.
Unicellular eukaryotic organisms are included in which kingdom?
Answer:
Unicellular eukaryotic organisms are included in kingdom Protista.

 

Question 44.
Give different types of Protists with examples.
Answer:
Protists are of different types:
(i) Plant like protists (Photosynthetic protists):
(a) They are termed as phytoplanktons, also known as Chrysophytes.
(b) They are autotrophic (photosynthetic) in nature and form major producers of ocean ecosystem.
(c)Most of them are referred as Diatoms because they have body wall made up of two soap-box like fitting silica covers. E.g. Diatoms.

1. Dinoflagellates:
(i) They are aquatic (mostly marine) and autotrophic (photosynthetic).
(ii) They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
(iii) The cell wall is made up of cellulosic stiff plates.
(iv) A pair of flagella is present, hence they are motile.
(v) They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

2. Euglenoids:
(i) They lack cell wall but have a tough covering of proteinaceous pellicle.
(ii) Pellicle covering provides flexibility and contractibility to Euglena.
(iii) They possess two flagella, one short and other long.
(iv) They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

(ii) Animal like protists (Consumer protists):
(a) They are the primitive animal forms.
(b) They are also termed as protozoans.
(c) These are heterotrophic and lack cell wall.
(d) Amoeboid protozoans have pseudopodia as locomotory organs. E.g. Amoeba, Entamoeba.
Amoeba is free living form, but Entamoeba is endoparasite and causes amoebic dysentery.
(e) Flagellated protozoans have flagella as locomotory organ. E.g. Trypanosoma.
(f) Cilliated protozoans have cilia for locomotion. E.g. Paramoecium.
(g) Plasmodium is a Sporozoan protozoa. It causes malaria. It forms spores in one of its life stages.

(iii) Fungi like protists (Consumer decomposer protists):
(a) They form a group called Myxomycetes.
(b) They are saprophytic in nature, found on decaying leaves.
(c) Their cells aggregate to form a large cell mass called plasmodium.
(d) The spores of plasmodium are very tough and survive extreme conditions, e.g. Slime molds.

Question 45.
Why diatoms are used in filtration and polishing?
Answer:
Diatoms forms a substance called Diatomaceous earth. These are the shells of diatoms containing silica that left behind for many years. Diatomaceous earth is granular; hence it is used in polishing and filtration.

Question 46.
Label the given figures representing ventral and dorsal view of Gonyaulax.

Answer:

Question 47.
Identify the following diagram, label it and write detail information in your words.
Answer:
The given figure represents Euglena.
Characteristics:
(i) It belongs to kingdom Protista. It is further classified into euglenoids.

1. Dinoflagellates:

1. They are aquatic (mostly marine) and autotrophic (photosynthetic).
2. They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
3. The cell wall is made up of cellulosic stiff plates.
4. A pair of flagella is present, hence they are motile.
5. They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

2. Euglenoids:

1. They lack cell wall but have a tough covering of proteinaceous pellicle.
2. Pellicle covering provides flexibility and contractibility to Euglena.
3. They possess two flagella, one short and other long.
4. They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

Question 48.
Write the characteristics of Kingdom plantae.
Answer:
Characteristics of Kingdom plantae:

1. Kingdom plantae is dominated by autotrophs.
2. Some members are insectivorous plants. E.g. Venus fly trap, pitcher plant, bladderwort, while some are heterotrophic parasitic members like Cuscuta.
3. Members of this kingdom are eukaryotic, multicellular, having eukaryotic cells containing chlorophyll.
4. Their cell wall is mostly made up of cellulose.
5. They exhibit alternation of generation i.e. life cycle has two distinct phases.
6. It is divided into two major groups Cryptogams and Phanerogams.

Question 49.
Give the general characters of Kingdom Fungi with examples.
Answer:
General characters of Kingdom Fungi:
1. Type of organisms: It is a unique kingdom of eukaryotic heterotrophic organisms, showing extracellular digestion. They may be unicellular or multicellular and filamentous. These are commonly found in warm and humid places.
2. Nucleus: The cells may be multinucleate or uninucleate.
3. Body: Multicellular organisms consist of a body called mycelium in which a number of thread or fibre-like structures called hyphae are present. The hyphae may be with septa (septate) or without septa (aseptate). The non-septate multinucleated hyphae are called coenocytic hyphae.
4. Cell wall: The cell wall in fungi is composed of chitin or fungal cellulose.
5. Cell organelles: The fungi contain well organized membrane bound cell organelles except the chloroplasts.
6. Nutrition: The fungi exhibit heterotrophic mode of nutrition and most of the members are saprophytes and absorb food which is decomposed (digested) outside. Some are parasitic or predators.
7. Reproduction: They reproduce both sexually as well as asexually. Asexual reproduction takes place by fragmentation, fission and budding.
8. Some fungi are symbiotic. These fungi either live with algae as lichens or as mycorrhiza in association with roots of higher plants.

Question 50.
Identify the following diagram, label it and write detail information in your words.
Answer:
The given figure represents Mucor.
Characteristics:

1. It belongs to class phycomycetes of kingdom fungi.
2. Mycelium is made up of aseptate coenocytic hyphae.
3. It commonly grows on decaying fruits,vegetables, in soil, on various food- stuff-like bread, jellies, jams, etc.
4. In favourable conditions mucor reproduces asexually by formation of spores within sporangia. It can also reproduce by sexual means.

Question 51.
Identify the following diagram, label it and write detail information in your words.
Answer:
The given figure represents Aspergillus.
Characteristics:

1. It belongs to class ascomycetes of kingdom Fungi.
2. It is multicellular.
3. The hyphae are branched and septate.
4. Aspergillus grows well in soil, decaying vegetation, hay, dung, on
5. plants, etc.
6. Asexual reproduction takes place by spores called conidia which are produced at the tip of hyphae called conidiophores.

Question 52.
Identify the following diagram, label it and write detail information in your words.
Answer:
The given figure represents Agaricus (Mushroom).
Characteristics:

1. It belongs to class basidiomycetes of kingdom Fungi.
2. It has branched septate hyphae.
3. It grows in soil, on rotten wood, etc.
4. It is edible and rich in proteins.
5. Vegetative reproduction takes place by fragmentation.

Question 53.
Explain how fungi exhibit heteromorphic mode of nutrition?
Answer:

1. Most of the members of kingdom fungi are saprophytes.
2. They absorb food which is decomposed (digested) outside.
3. Some are parasites or predators and some are symbiotic.
4. In fungi, chloroplast is absent, thus they cannot synthesize their own food by photosynthesis. Due to this, fungi exhibit heteromorphic mode of nutrition.

Quesiton 54.
Identify the given picture and explain in detail.
Answer:
The given picture represents Lichens.

1. Lichen is an association of an alga and fungus.
2. It is the best example of symbiosis or mutualism.
3. They are found in extreme environments like snow clad poles.
4. The algal component of lichen is phycobiont, mostly belongs to cyanobacteria (blue-green algae) or green algae and fungal component is mycobiont.
5. Algae prepares the food and supplies it to the fungal component, while fungal component provides shelter to algae and also absorbs water and minerals for algae.
6. The association is intense and it is difficult to identify them as separate living beings.
7. They are very sensitive to pollutions, hence not found in polluted areas.
8. They are used as pollution indicators.
9. They play an important role in soil formation by using specific acid productions.
   [Note: Lichens cannot be categorized as acellular organisms]

Question 55.
Write the general characters of Kingdom Animalia with examples.
Answer:
General characters of Kingdom Animalia:

1. Types of organisms: The organisms are multicellular and eukaryotic.
2. Habitat: The organisms may be aquatic, terrestrial, amphibious or aerial in habitat.
3. Cell organelles: The organisms do not possess cell wall, plastids and central vacuole.
4. Locomotion: Majority of the animals are motile. However, few like sponges are sedentary.
5. Sense orgAnswer: They possess sense organs, nervous system and respond to stimuli by exhibiting certain behaviour.
6. Reproduction: They mostly reproduce sexually by producing gametes, while some can reproduce asexually.
7. Nutrition: They are heterotrophic, mostly holozoic, sometimes parasitic.
8. Growth: It is determinate, (follow definite pattern)

Question 56.
Observe and discuss:
Complete the following table on the basis of previous knowledge.

Characters	Monera	Protista	Fungi	Plantae	Animalia
Cell type	Prokaryotic	Eukaryotic	Eukaryotic	Eukaryotic	Eukaryotic
Cell wall		Present in some organisms		Present (cellulose)
Nuclear membrane	Absent	Present	Present		Present
Body organization	Unicellular		Multicellular/ loose tissue	Tissue /organ	Tissue            /organ system
Mode of nutrition		Autotrophic Photosynthetic, Heterotrophic		Autotrophic (Photosynthetic)
Ecological role	Decomposers		Decomposers		Consumers

Answer:

Characters	Monera	Protista	Fungi	Plantae	Animalia
Cell type	Prokaryotic	Eukaryotic	Eukaryotic	Eukaryotic	Eukaryotic
Cell wall	Present (Peptidoglycan)	Present in some organisms	Present (chitin)	Present (cellulose)	Absent
Nuclear membrane	Absent	Present	Present	Present	Present
Body organization	Unicellular	Unicellular	Multicellular/ loose tissue	Tissue /organ	Tissue  /organ system
Mode of nutrition	Heterotrophic (saprophytic/ parasitic) Autotrophic (Photoautotrophic/ Chemoautotrophic)	Autotrophic Photosynthetic, Heterotrophic	Heterotrophic (saprophytic/ parasitic)	Autotrophic (Photosynthetic)	Heterotrophic (holozoic)
Ecological role	Decomposers	Producers and consumers	Decomposers	Producers	Consumers

Question 57.
Who referred virus as ‘contagium vivum fluidum’?
Answer:
M. W. Beijerinck referred virus as ‘contagium vivum fluidum (infectious living fluid).’

Question 58.
Who demonstrated that viruses are inert outside the host cell and can be crystallised?
Answer:
Stanley demonstrated that viruses are inert and can be crystallised.
[Note: Students can scan the adjacent QR code for detail classification of given tree diagram.]

Question 59.
What is the structure of virus?
Answer:

1. Viruses are acellular and ultramicroscopic.
2. The genetic material in viruses is either single or double-stranded RNA or double-stranded DNA.
3. Their genetic material is protected by a protein coat called capsid.
4. Capsid is made up of smaller units called capsomeres.
5. Capsomeres are arranged in polyhedral or helical forms thus, imparting that particular shape to the virus.

Question 60.
Give examples of:
1. Diseases caused by viruses in plants:
2. Diseases caused by viruses in animals:
Answer:
1. Diseases caused by viruses in plants: Leaf curling, yellowing, mosaic formation, etc.
2. Diseases caused by viruses in animals: Swine flu, Small pox, mumps, herpes, common cold, AIDS, etc.

Question 61.
Write a short note on viroids.
Answer:
Viroids:

1. These are mainly plant pathogens.
2. Viroids were discovered by Theodor Diener.
3. The first viroid discovered was PSTV (Potato spindle tuber viroid) which causes a disease in potato.
4. Viroids are very small, circular, single stranded RNA which are without any protein coat.
5. Viroids are smaller in size than viruses.

Question 62.
Apply Your Knowledge:

Question 1.
In your laboratory you accidentally discover an old permanent slide without a label. You are curious to identify it, and you place the slide under the microscope. You observe the following features:
1. Well-organized nucleus
2. Unicellular
3. Biflagellate – one placed longitudinally and the other transversely.
Answer:
All unicellular eukaryotes form a connecting link between prokaryotic Kingdom Monera and complex eukaryotic Kingdoms Plantae, Fungi and Animalia. Since the specimen shows the presence of two flagella, one placed longitudinally and the other transversely, the given organism can be dinoflagellate and has to be placed under Kingdom Protista.

Question 2.
Name the following:
1. The kingdom which includes the smallest living forms.
2. The protists which behave as heterotroph in absence of light but performs photosynthesis in presence of light
3. These are infectious single stranded RNA, smaller than virus
Answer:
1. Kingdom Monera
2. Euglena
3. Viroids

Question 63.
Quick Review
Answer:
Taxonomic Hierarchy

Kingdom → Sub-kingdom → Division/phylum → Class → Cohort /order → Family → Genus → Species

Question 64.
Exercise

Question 1.
Define Systematics.
Answer:
Systematics is the study of kinds and diversity of organisms and their comparative and evolutionary relationship.

Question 2.
What is classification?
Answer:
Classification is the arrangement of organisms or groups of organisms in distinct categories in accordance with a particular and well-established plan.

Question 3.
Explain different methods of classification.
Answer:
The three types of classification systems are:
(i) Artificial system:
(a) It is based on few visible, easily observable characters, which are non-evolutionary such as habit, colour, form, etc.
(b) It does not consider the affinities (relationships) among different organisms.
E.g. Linnaeus system of classification.

(ii) Natural system:
It is based on objectively significant characters with respect to their affinities with other organisms.
E.g. Bentham and Hooker’s system of classification.

(iii) Phylogenetic system:
It is based on the phylogenetic relationship between different organisms with respect to common evolutionary descent (ancestor).
E.g. Engler and Prantl’s classification.

Question 4.
Domain eukarya has which cells?
Answer:
Domain Eukarya has eukaryotic cells.

Question 5.
Name three domains of life.
Answer:
Three domains of life are Archaea, Bacteria and Eukarya.

Question 6.
Write a short note on chemotaxonomy.
Answer:
1. It is method of biological classification based on the similarities and differences in structure of certain chemical compounds present among the organisms being classified.
2. Thus, it is a classification based on chemical constituents of organisms.
3. For e.g. Cell wall with peptidoglycan is present in Bacteria while it is absent in Archaea.
Among Eukarya, fungi have chitinous cell wall, while plants have cellulosic cell wall.

Question 7.
What is numerical taxonomy? Who proposed it?
Answer:
Numerical taxonomy:

1. It is based on quantification of characters and develops an algorithm for classification.
2. The aim of this was to create a taxonomy using numeric algorithms like cluster analysis rather than using subjective evaluation of their properties.
3. It was proposed by Sokel and Sneath in 1963.

Question 8.
Write a note on cladogram?
Answer:
1. It is a representation of hypothetical relationship denoting a comparison of organisms and their common ancestors.
2. It has a typical branching pattern.

Question 9.
Write a short note on phylogeny.
Answer:
Phylogeny:

1. It is the evolutionary relationship of organism.
2. It is an important tool in classification as it considers not merely the morphological status but also the relationship of one group of organisms with other groups of life.
3. The system helps to understand the evolution and also focuses on the similarities of their metabolic functioning.
4. Woese’s three domain concept, as well as Whittaker’s five-kingdom system, are examples of phylogenetic relationship.

Question 10.
Explain DNA barcoding.
Answer:
DNA barcoding is a new method for identification of any species based on its DNA sequence, which is obtained from a tiny tissue sample of the organism under study.
DNA barcoding helps to study newly identified species as well as understanding ecological and evolutionary relationships between living organisms.
The process of DNA barcoding includes two basic steps:
(i) Collecting DNA barcode data of known species.
(ii) Matching the barcode sequence of the unknown sample against the barcode library for identification.

The applications of DNA barcoding are as follows:

1. It helps to protect endangered species.
2. It plays an important role in preservation of natural resources.
3. It is also used for pest control in agriculture.
4. It is used for identification of disease vectors.
5. It is used for authentication of natural health products.
6. It is also used for identification of medicinal plants.

Question 11.
Explain the term taxonomic category.
Answer:

1. Category is a rank or level in the hierarchial classification of organisms.
2. Each category is referred to as a unit of classification.
3. Category is a part of taxonomic arrangements hence, called taxonomic category.
4. All categories together constitute the taxonomic hierarchy.

Question 12.
Give the classification of cobra.
Answer:
Cobra

Question 13.
Give the classification of china-rose.
Answer:
China Rose

Question 14.
What is taxon? Give any one example of it.
Answer:
1. Taxon is a group of living organisms of any rank in the system of classification.
2. In plant kingdom, each taxonomic group such as angiospermae, dicotyledonae, polypetalae, malvaceae represents a taxon.

Question 15.
Which are the units of classification?
Answer:
(i) Species:
(a) Species is the principal natural taxonomic unit, ranking below a genus.
(b) It is a group of organisms that can interbreed under natural condition to produce fertile offspring.
(c) It was thought to be an indivisible, stable and static unit.
(d) However, in the modem taxonomy, subdivision of species such as sub-species, varities and populations are seen and given more importance.

(ii) Genus:

(a) Genus is a taxonomic rank or category larger than species used in the biological classification of living and fossil organisms.
(b) Genus is a group of species bearing close resemblance to one another in their morphological characters but they do not interbreed.
(c) For e.g. Tiger, Leopard, Lion all three belong to same genus Panthera. They have common characters yet are different from each other because their genus is same but species is different.
(d) Another example is genus Solarium. Brinjal and potato both belong to this genus.

(iii) Family:

(a) It is one of the major hierarchial taxonomic rank.
(b) A family represents a group of closely related genera.
(c) For e.g. genera like Hibiscus, Gossypium, Sida, Bombax are included in same family Malvaceae.
(d) Although, there are many similarities between cat and dog, cat belongs to the family of leopards, tigers and lions, i.e. family Felidae and dog belongs to different family i.e. Canidae.

(iv) Cohort/Order:

(a) It is taxonomic rank used in the classification of organisms and recognised by nomenclature codes.
(b) An order is a group of closely related families showing definite affinities.
(c) Members belonging to same order but different families may show very few dissimilarities.
(d) For e.g. family Papaveraceae, Brassicaceae, Capparidaceae, etc with parietal placentation are grouped in order Parietales.
(e) Families of dogs and cats though are different, they belong to same order Carnivora.

(v) Class:

(a) The class is the distinct taxonomic rank of biological classification having its own distinctive name.
(b) Class is the assemblage of closely allied orders.
(c) For e.g. Orders Carnivora and order Primates belong to class Mammalia. Thus monkeys, gorillas, gibbons (Primates) and dogs, cats, tigers (Carnivora) belong to same class.

(vi) Division/ Phylum:

(a) The division is a category composed of related classes.
(b) For e.g. division Angiospermae includes two classes Dicotyledonae and Monocotyledonae.
(c) In animal classification, instead of division, the category Phylum is used.

(vii) Sub-kingdom:

(a) Different divisions having some similarities form sub-kingdom.
(b) The divisions Angiospermae and Gymnospermae forms the sub-kingdom Phanerogams or Spermatophyta (all seed producing plants).

(viii) Kingdom:

(a) It is the highest taxonomic category composed of different sub-kingdoms.
(b) For e.g. sub-kingdom Phanerogams and Cryptogams form the Plant kingdom or Plantae which includes all the plants, while all animals are included in kingdom Animalia.

Question 16.
Explain the following terms by giving one example of each:
1. Sub-kingdom
2. Genus
3. Order
Answer:
1. Sub-kingdom:

(a) Different divisions having some similarities form sub-kingdom.
(b) The divisions Angiospermae and Gymnospermae forms the sub-kingdom Phanerogams or Spermatophyta (all seed producing plants).

2. Genus:

(a) Genus is a taxonomic rank or category larger than species used in the biological classification of living and fossil organisms.
(b) Genus is a group of species bearing close resemblance to one another in their morphological characters but they do not interbreed.
(c) For e.g. Tiger, Leopard, Lion all three belong to same genus Panthera. They have common characters yet are different from each other because their genus is same but species is different.
(d) Another example is genus Solarium. Brinjal and potato both belong to this genus.

3. Cohort/Order:

(a) It is taxonomic rank used in the classification of organisms and recognised by nomenclature codes.
(b) An order is a group of closely related families showing definite affinities.
(c) Members belonging to same order but different families may show very few dissimilarities.
(d) For e.g. family Papaveraceae, Brassicaceae, Capparidaceae, etc with parietal placentation are grouped in order Parietales.
(e) Families of dogs and cats though are different, they belong to same order Carnivora.

Question 17.
‘A family represents a group of closely related genera’. Give one example to justify the statement.
Answer:
(c) For e.g. genera like Hibiscus, Gossypium, Sida, Bombax are included in same family Malvaceae.

Question 18.
What does letter ‘L’ indicates in Man gif era indica L., ?
Answer:
In Mangifera indica L., letter L indicates author’s name i.e. Linnaeus.

Question 19.
1. Define binomial nomenclature system.
2. Who proposed it?
3. Why a unique name for a particular individual is essential in a multilingual country like India?
Answer:
1. The name of the organism is composed of two Latin or Greek words.
2. Generic epithet is a simple noun which should come first and always begin with a capital letter.
3. The generic and specific epithet must be underlined separately if hand written or in italics when printed.

Question 20.
Why is binomial nomenclature useful for classification of organisms?
Answer:
Binomial nomenclature is important because:

1. The binomials are simple, meaningful and precise.
2. They are standard since they do not change from place to place.
3. These names avoid confusion and uncertainty created by local or vernacular names. The organisms are known by the same name throughout the world.
4. The binomials are easy to understand and remember.
5. It indicates phylogeny (evolutionary history) of organisms.
6. It helps to understand inter-relationship between organisms.

Question 21.
Which are the two kingdoms of organisms given by Carl Linnaeus? What was the drawback of this system?
Answer:
The two-kingdom system of classification included Kingdom plantae and Kingdom animalia. This system was introduced by Carl Linnaeus. Two kingdom system was found inadequate for classification of some organisms like bacteria, fungi, Euglena, etc.

Question 22.
Name the five kingdoms given by Whittaker?
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

1. Kingdom Monera
2. Kingdom Protista
3. Kingdom Plantae
4. Kingdom Fungi
5. Kingdom Animalia

Question 23.
Unicellular prokaryotic organisms are included in which kingdom?
Answer:
(i) Size: The organisms included in this kingdom are microscopic, unicellular and prokaryotic.

Question 24.
Explain kingdom Monera with the help of given points:
i. Nucleus
ii. Reproduction
iii. Nutrition
Answer:
(i) Nucleus: These organisms do not have well defined nucleus. DNA exists as a simple double stranded circular single chromosome called as nucleoid. Apart from the nucleoid they often show presence of extrachromosomal DNA which is small circular called plasmids.
(ii) Reproduction: The mode of reproduction is asexual or with the help of binary fission or budding. Very rarely, sexual reproduction occurs by conjugation method.
(iii) Nutrition: Majority are heterotrophic, parasitic or saprophytic in nutrition. Few are autotrophic that can be either photoautotrophs or chemoautotrophs.

Question 25.
Give examples of archaebacteria and eubacteria.
Answer:
Examples:
Archaebacteria: e.g. Methanobacillus, Thiobacillus, etc.
Eubacteria: e.g. Chlorobium, Chromatium, and Cyanobacteria e.g. Nostoc, Azotobacter, etc.

Question 26.
What is mycoplasma?
Answer:
1. These are the smallest living cells known.
2. They lack cell wall.
3. Many forms are pathogenic.
4. They are resistant to common antibiotics because they lack cell wall.

Question 27.
Enlist different types of protozoa.
Answer:
(ii) Animal like protists (Consumer protists):
(a) They are the primitive animal forms.
(b) They are also termed as protozoans.
(c) These are heterotrophic and lack cell wall.
(d) Amoeboid protozoans have pseudopodia as locomotory organs. E.g. Amoeba, Entamoeba.
Amoeba is free living form, but Entamoeba is endoparasite and causes amoebic dysentery.
(e) Flagellated protozoans have flagella as locomotory organ. E.g. Trypanosoma.
(f) Cilliated protozoans have cilia for locomotion. E.g. Paramoecium.
(g) Plasmodium is a Sporozoan protozoa. It causes malaria. It forms spores in one of its life stages.

Question 28.
Which are the different types of protists?
Answer:
Protists are of different types:
(i) Plant like protists (Photosynthetic protists):
(a) They are termed as phytoplanktons, also known as Chrysophytes.
(b) They are autotrophic (photosynthetic) in nature and form major producers of ocean ecosystem.
(c)Most of them are referred as Diatoms because they have body wall made up of two soap-box like fitting silica covers. E.g. Diatoms.

1. Dinoflagellates:
(i) They are aquatic (mostly marine) and autotrophic (photosynthetic).
(ii) They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
(iii) The cell wall is made up of cellulosic stiff plates.
(iv) A pair of flagella is present, hence they are motile.
(v) They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

2. Euglenoids:
(i) They lack cell wall but have a tough covering of proteinaceous pellicle.
(ii) Pellicle covering provides flexibility and contractibility to Euglena.
(iii) They possess two flagella, one short and other long.
(iv) They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

(ii) Animal like protists (Consumer protists):
(a) They are the primitive animal forms.
(b) They are also termed as protozoans.
(c) These are heterotrophic and lack cell wall.
(d) Amoeboid protozoans have pseudopodia as locomotory organs. E.g. Amoeba, Entamoeba.
Amoeba is free living form, but Entamoeba is endoparasite and causes amoebic dysentery.
(e) Flagellated protozoans have flagella as locomotory organ. E.g. Trypanosoma.
(f) Cilliated protozoans have cilia for locomotion. E.g. Paramoecium.
(g) Plasmodium is a Sporozoan protozoa. It causes malaria. It forms spores in one of its life stages.

(iii) Fungi like protists (Consumer decomposer protists):
(a) They form a group called Myxomycetes.
(b) They are saprophytic in nature, found on decaying leaves.
(c) Their cells aggregate to form a large cell mass called plasmodium.
(d) The spores of plasmodium are very tough and survive extreme conditions, e.g. Slime molds.

Question 29.
What are dinoflagellates?
Answer:
Protists are of different types:
(i) Plant like protists (Photosynthetic protists):
(a) They are termed as phytoplanktons, also known as Chrysophytes.
(b) They are autotrophic (photosynthetic) in nature and form major producers of ocean ecosystem.
(c)Most of them are referred as Diatoms because they have body wall made up of two soap-box like fitting silica covers. E.g. Diatoms.

1. Dinoflagellates:
(i) They are aquatic (mostly marine) and autotrophic (photosynthetic).
(ii) They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
(iii) The cell wall is made up of cellulosic stiff plates.
(iv) A pair of flagella is present, hence they are motile.
(v) They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

Question 30.
Explain animal like protists.
Answer:
(ii) Animal like protists (Consumer protists):
(a) They are the primitive animal forms.
(b) They are also termed as protozoans.
(c) These are heterotrophic and lack cell wall.
(d) Amoeboid protozoans have pseudopodia as locomotory organs. E.g. Amoeba, Entamoeba.
Amoeba is free living form, but Entamoeba is endoparasite and causes amoebic dysentery.
(e) Flagellated protozoans have flagella as locomotory organ. E.g. Trypanosoma.
(f) Cilliated protozoans have cilia for locomotion. E.g. Paramoecium.
(g) Plasmodium is a Sporozoan protozoa. It causes malaria. It forms spores in one of its life stages.

Question 31.
Give examples of insectivorous plants.
Answer:
(ii) Some members are insectivorous plants. E.g. Venus fly trap, pitcher plant, bladderwort, while some are heterotrophic parasitic members like Cuscuta.

Question 32.
What are the two major group in which kingdom plantae is divided?
Answer:
(vi) It is divided into two major groups Cryptogams and Phanerogams.

Question 33.
Explain fungi like protist.
Answer:
(iii) Fungi like protists (Consumer decomposer protists):
(a) They form a group called Myxomycetes.
(b) They are saprophytic in nature, found on decaying leaves.
(c) Their cells aggregate to form a large cell mass called plasmodium.
(d) The spores of plasmodium are very tough and survive extreme conditions, e.g. Slime molds.

Question 34.
What are the characteristics of euglenoids?
Answer:
Euglenoids:

1. They lack cell wall but have a tough covering of proteinaceous pellicle.
2. Pellicle covering provides flexibility and contractibility to Euglena.
3. They possess two flagella, one short and other long.
4. They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

Question 35.
Explain in detail general characters of Kingdom Fungi.
Answer:
Euglenoids:

1. They lack cell wall but have a tough covering of proteinaceous pellicle.
2. Pellicle covering provides flexibility and contractibility to Euglena.
3. They possess two flagella, one short and other long.
4. They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

Question 36.
Why do fungi exhibit heterotrophic mode of nutrition?
Answer:
Nutrition: The fungi exhibit heterotrophic mode of nutrition and most of the members are saprophytes and absorb food which is decomposed (digested) outside. Some are parasitic or predators.

Question 37.
Name the four classes of kingdom fungi.
Answer:
Fungi are classified into four types on the basis of their structure, mode of spore formation and fruiting bodies as follows:
1. Phycomycetes:
Members of this class are commonly called as algal fungi.
These are consisting of aseptate coenocytic hyphae.
They grow well in moist and damp places on decaying organic matter as well as in aquatic habitats or as parasites on plants.
e.g. Mucor, Rhizopus (bread mold), Albugo (parasitic fungus on mustard).

2. Ascomycetes:
These are commonly called as sac fungi.
These are multicellular. Rarely they are unicellular (e.g. Yeast).
Hyphae are branched and septate.
They can be decomposers, parasites or coprophilous (grow on dung).
Some varieties of this class are consumed as delicacies such as morels and truffles.
Neurospora is useful in genetic and biochemical assays.
e.g. Aspergillus, Penicillium, Neurospora, Claviceps, Saccharomyces (unicellular ascomycetes).

3. Basidiomycetes:
These are commonly called as club fungi.
They have branched septate hyphae.
e.g. Agaricus (mushrooms), Ganoderma (bracket fungi), Ustilago (smuts), Puccinia (rusts), etc.

4. Deuteromycetes:
It is a group of fungi which are known to reproduce only asexually.
They are commonly called imperfect fungi.
They are mainly decomposers, while few are parasitic, e.g. Alternaria.

Question 38.
Explain in detail the class of kingdom fungi which includes yeast.
Answer:
Ascomycetes:

1. These are commonly called as sac fungi.
2. These are multicellular. Rarely they are unicellular (e.g. Yeast).
3. Hyphae are branched and septate.
4. They can be decomposers, parasites or coprophilous (grow on dung).
5. Some varieties of this class are consumed as delicacies such as morels and truffles.
6. Neurospora is useful in genetic and biochemical assays.
   e.g. Aspergillus, Penicillium, Neurospora, Claviceps, Saccharomyces (unicellular ascomycetes).

Question 39.
Why deuteromycetes are called imperfect fungi?
Answer:
Deuteromycetes:

• It is a group of fungi which are known to reproduce only asexually.
• They are commonly called imperfect fungi.
• They are mainly decomposers, while few are parasitic, e.g. Alternaria.

Question 40.
What are lichens?
Answer:
The given picture represents Lichens.

1. Lichen is an association of an alga and fungus.
2. It is the best example of symbiosis or mutualism.
3. They are found in extreme environments like snow clad poles.
4. The algal component of lichen is phycobiont, mostly belongs to cyanobacteria (blue-green algae) or green algae and fungal component is mycobiont.
5. Algae prepares the food and supplies it to the fungal component, while fungal component provides shelter to algae and also absorbs water and minerals for algae.
6. The association is intense and it is difficult to identify them as separate living beings.
7. They are very sensitive to pollutions, hence not found in polluted areas.
8. They are used as pollution indicators.
9. They play an important role in soil formation by using specific acid productions.
   [Note: Lichens cannot be categorized as acellular organisms]

Question 41.
What is the fungal partner in lichen called?
Answer:
The algal component of lichen is phycobiont, mostly belongs to cyanobacteria (blue-green algae) or green algae and fungal component is mycobiont.

Question 42.
What is the algal partner in lichen called?
Answer:
The algal component of lichen is phycobiont, mostly belongs to cyanobacteria (blue-green algae) or green algae and fungal component is mycobiont.

Question 43.
Why lichens are considered as pollution indicators?
Answer:
They are very sensitive to pollutions, hence not found in polluted areas.

Question 44.
Holozoic mode of nutrition is observed in which kingdom?
Answer:
Nutrition: They are heterotrophic, mostly holozoic, sometimes parasitic.

Question 45.
Who coined the name contagium vivum fluidum?
Answer:
M. W. Beijerinck referred virus as ‘contagium vivum fluidum (infectious living fluid).’

Question 46.
What is the genetic material in viruses?
Answer:
Viruses possess their own genetic material in the form of either DNA or RNA, but never both. The genetic material in viruses is covered by a protein coat (capsid), hence called nucleoprotein.

Question 47.
What are bacteriophages?
Answer:
Bacteriophage:
(a) They have tadpole-like shape.
(b) They infect bacteria and hence are called as bacteriophage.
(c) Bacteriophages were discovered by Twort.
(d) Bacteriophages have double stranded DNA as the genetic material.
(e) Its body consists of head, collar and tail.

Question 48.
Give example of viral disease caused in humans.
Answer:
Diseases caused by viruses in animals: Swine flu, Small pox, mumps, herpes, common cold, AIDS, etc.

Question 49.
Who discovered viroids?
Answer:
Viroids were discovered by Theodor Diener.

Question 50.
What is the genetic material in viroids?
Answer:
Viroids are very small, circular, single stranded RNA which are without any protein coat.

Question 51.
Multiplechoice Questions:

Question 1.
The term ‘Taxonomy’ was coined by
(A) Carl Linnaeus
(B) A.P. de Candolle
(C) Carl Woese
(D) R.H Whittaker
Answer:
(B) A.P. de Candolle

Question 2.
Arrangement of organisms into distinct categories is called
(A) Taxonomy
(B) Taxon
(C) Nomenclature
(D) Classification
Answer:
(D) Classification

Question 3.
The domain known for its survival in very extreme condition like high temperature, salinity, etc. is
(A) Eukarya
(B) Archaea
(C) Bacteria
(D) Cyanobacteria
Answer:
(B) Archaea

Question 4.
Kingdom Protista, Fungi, Plantae and Animalia are included under domain
(A) Eukarya
(B) Archaea
(C) Bacteria
(D) Cyanobacteria
Answer:
(A) Eukarya

Question 5.
Which system of classification was based upon easily observable characters?
(A) Natural
(B) Phylogenetic
(C) Artificial
(D) DNA barcoding
Answer:
(C) Artificial

Question 6.
System based upon chemical constituents of organisms is
(A) Cladogram
(B) Phylogeny
(C) DNA barcoding
(D) chemotaxonomy
Answer:
(D) chemotaxonomy

Question 7.
Woese’s three domain and Whittaker’s five kingdom concept is based upon
(A) Visible characters
(B) Phylogenetic relationship
(C) Numerical taxonomy
(D) DNA barcoding
Answer:
(B) Phylogenetic relationship

Question 8.
A taxonomic group of any rank is called
(A) grade
(B) category
(C) variety
(D) taxon

Question 9.
One of the following has correct descending sequence hierarchy
(A) class, division, order, family
(B) division, class, order, family
(C) order, family, class, division
(D) family, order, class, genus
Answer:
(B) division, class, order, family

Question 10.
Which among the following is an order?
(A) Malvales
(B) Polypetalae
(C) Angiospermae
(D) Hibiscus
Answer:
(A) Malvales

Question 11.
The basic unit of classification
(A) genus
(B) species
(C) kingdom
(D) family
Answer:
(B) species

Question 12.
As we go higher in taxonomical ladder i.e. from species to kingdom, the number of common characters
(A) remains constant
(B) goes on increasing
(C) goes on decreasing
(D) increases till class and then starts decreasing
Answer:
(C) goes on decreasing

Question 13.
Group of species which resemble closely in morphological characters but do not interbreed is called
(A) genus
(B) species
(C) family
(D) order
Answer:
(A) genus

Question 14.
Highest category of taxonomy is
(A) species
(B) class
(C) order
(D) kingdom
Answer:
(D) kingdom

Question 15.
Carl Linnaeus introduced binomial system of nomenclature in his book
(A) Species Plantarum
(B) ICBN
(C) Plantarum Linnaeus
(D) Species Linnaeus
Answer:
(A) Species Plantarum

Question 16.
Before 2011, scientific names were confirmed by
(A) ICBN
(B) IBC
(C) ICZN
(D) IBA
Answer:
(A) ICBN

Question 17.
Which code is also known as “Shenzhen code”?
(A) ICBN
(B) IBC
(C) ICZN
(D) IBA
Answer:
(B) IBC

Question 18.
In Helianthus annuus, ‘annuus ’ indicates
(A) genus
(B) species
(C) family
(D) class
Answer:
(B) species

Question 19.
In five kingdom classification, unicellular prokaryotes are included in kingdom
(A) Protista
(B) Fungi
(C) Monera
(D) Animalia
Answer:
(C) Monera

Question 20.
The bacteria that can withstand high salinities are called
(A) Saltophiles
(B) Thermophiles
(C) Halophiles
(D) Psychrophiles
Answer:
(C) Halophiles

Question 21.
The bacteria that can withstand extreme temperature are known as
(A) Saltophiles
(B) thermophiles
(C) Halophiles
(D) both (A) and (C)
Answer:
(B) thermophiles

Question 22.
Bacillus is
(A) comma shaped
(B) rod shaped
(C) kidney shaped
(D) spiral
Answer:
(B) rod shaped

Question 23.
Which organism belongs Monera?
(A) Cyanobacteria
(B) Mushroom
(C) Euglena
(D) Moss
Answer:
(A) Cyanobacteria

Question 24.
_________ is an example of plant like protists.
(A) Diatoms
(B) Ustilago
(C) Entamoeba
(D) Euglena
Answer:
(A) Diatoms

Question 25.
Fungi like protist are also called as ________.
(A) Myxomycetes
(B) Mycomycetes
(C) Mycoplasm
(D) Yeast
Answer:
(A) Myxomycetes

Quesiton 26.
The body of a fungus is made up of
(A) hyphae
(B) sporangium
(C) rhizoid
(D) fruiting body
Answer:
(A) hyphae

Question 27.
Agaricus belongs to class
(A) Deuteromycetes
(B) Phycomycetes
(C) Basidiomycetes
(D) Ascomycetes
Answer:
(C) Basidiomycetes

Question 28.
Which of the following is harmful fungus that causes diseases in plants?
(A) Puccinia
(B) Mushroom
(C) Yeast
(D) Streptomyces
Answer:
(A) Puccinia

Question 29.
Which of the following is NOT true about kingdom animalia?
(A) Members are heterotrophs.
(B) They lack chlorophyll as well as cell wall.
(C) Growth is indeterminate.
(D) Most of the members have capacity of locomotion.
Answer:
(C) Growth is indeterminate.

Question 30.
Which of the following are virus free varieties of banana produced by tissue culture technique?
(A) Shrimanti
(B) Basarai
(C) G-9
(D) All of these
Answer:
(D) All of these

Question 31.
The fungal component of lichen is called
(A) phycobiont
(B) photobiont
(C) mycobiont
(D) symbiont
Answer:
(C) mycobiont

Question 52.
Competitive Corner:

Question 1.
Which of the following is against the rules of ICBN?
(A) Generic and specific names should be written starting with small letters.
(B) Hand written scientific names should be underlined.
(C) Every species should have a generic name and a specific epithet.
(D) Scientific names are in Latin and should be italized.
Hint: The generic name should start with a capital letter while the species name should start with small letter.
Answer:
(A) Generic and specific names should be written starting with small letters.

Question 2.
Select the correctly written scientific name of Mango which was first described by Carolus Linnaeus: [NEET Odisha 2019]
(A) Mangifera indica
(B) Mangifera Indica
(C) Mangifera indica Car. Linn.
(D) Mangifera indica Linn
Hint: The author’s name appears after the specific epithet i.e. at the end of the biological name in this manner – Mangifera indica Linn.
Answer:
(D) Mangifera indica Linn

Question 3.
Match the organisms in Column I with habitats in Column II.

Column I	Column II
1. Halophiles	(a) Hot springs
2. Thermoacidophiles	(b) Aquatic environment
3. Methanogens	(c) Guts of ruminants
4. Cyanobacteria	(d) Salty areas

Select the correct answer from the options given below:
(A) i – b, ii – d, iii – c, iv – a
(B) i – d, ii – a, iii – c, iv – b
(C) i – a, ii – b, iii – c, iv – d
(D) i – c, ii – d, iii – b, iv – a
Answer:
(B) i – d, ii – a, iii – c, iv – b

Question 4.
Which of the following statements is CORRECT?
(A) Lichens are not good pollution indicators.
(B) Lichens do not grow in polluted areas.
(C) Algal component of lichens is called mycobiont.,
(D) Fungal component of lichens is called phycobiont.
Hint: Lichens bare good pollution indicators as they do not grow in polluted areas.
Answer:
(B) Lichens do not grow in polluted areas.

Question 5.
Match Column – I with Column – II. Choose the correct answer from the options given below:

Column – I	Column – II
1. Saprophyte	(a) Symbiotic association of fungi with plants roots
2. Parasite	(b) Decomposition of dead
3. Lichens	(c) Living on living plants or animals
4. Mycorrhiza	(d) Symbiotic association of algae and fungi

(A) i – q, ii – p, iii – r, iv – s
(B) i – q, ii – r, iii – s, iv – p
(C) i – p, ii – q, iii – r, iv – s
(D) i – r, ii – q, iii – p, iv – s
Answer:
(B) i – q, ii – r, iii – s, iv – p

Question 6.
Lowest category in the hierarchial system of classification is
(A) species
(B) order
(C) kingdom
(D) genus
Answer:
(A) species

Question 7.
Which group of fungi is called imperfect fungi?
(A) Ascomycetes
(B) Phycomycetes
(C) Deuteromycetes
(D) Basidiomycetes
Answer:
(C) Deuteromycetes

Question 8.
Which one of the following is an Incorrect pair?
(A) Three kingdom system of classification → Haeckel
(B) Three domain system of classification → Adolf Mayr
(C) Five Kingdom system of classification → R.H.Whittaker
(D) Two kingdom system of classification → Carolus Linnaeus
Hint: Three domain system of classification → Carl Woese
Answer:
(B) Three domain system of classification → Adolf Mayr

Question 9.
In the system of classification, which one of the following is NOT a category?
(A) Kingdom
(B) Series
(C) Angiospermae
(D) Genus
Hint: Angiospermae is a taxon.
Answer:
(C) Angiospermae

Question 10.
Which one of the following characteristics is NOT shown by a virus?
(A) They are acellular.
(B) They can be crystallised.
(C) Active outside the host’s body.
(D) Have genetic material.
Hint: Viruses are inert outside the host cell.
Answer:
(C) Active outside the host’s body.

Question 11.
Select the WRONG statement.
(A) Pseudopodia are locomotory and feeding structures in Sporozoans.
(B) Mushrooms belong to Basidiomycetes.
(C) Cell wall is present in members of Fungi and Plantae.
(D) Mitochondria are the powerhouse of the cell in all kingdoms except Monera.
Hint: Pseudopodia are locomotory and feeding structures in Protozoans.
Answer:
(A) Pseudopodia are locomotory and feeding structures in Sporozoans.

Question 12.
Which of the following are found in extreme saline conditions?
(A) Archaebacteria
(B) Eubacteria
(C) Cyanobacteria
(D) Mycobacteria
Hint: Bacteria found in extremely saline conditions are called halophiles. Archaebacteria includes bacteria that survive in most harsh habitats such as extreme salty area, hot springs and marshy area.
Answer:
(A) Archaebacteria

Question 13.
Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen?
(A) Bacillus
(B) Pseudomonas
(C) Mycoplasma
(D) Nostoc

Question 14.
Viroids differ from viruses in having
(A) DNA molecules with protein coat
(B) DNA molecules without protein coat
(C) RNA molecules with protein coat
(D) RNA molecules without protein coat
Hint: Viroids are smaller than viruses. They are regarded as sub-viral agents or free RNA, without protein coat (usually found in viruses). They are infectious RNA. e.g. Potato spindle tuber disease.
Answer:
(D) RNA molecules without protein coat