Bhagya

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

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 7 Company Meetings – I Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 7 Company Meetings – I

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

Question 1.
The business to be transacted at the meeting is called _____________
(a) Agenda
(b) Minutes
(c) Notice
Answer:
(a) Agenda

Question 2.
A proxy form must be deposited in the registered office of the company at least _____________ hours before the meeting.
(a) 48
(b) 24
(c) 12
Answer:
(a) 48

Question 3.
A copy of the special resolution must be filed with the registrar within _____________ days of the passing resolution.
(a) 30
(b) 60
(c) 90
Answer:
(a) 30

Question 4.
Quorum for a private company is _____________ members.
(a) 15
(b) 5
(c) 2
Answer:
(c) 2

Question 5.
Point of order can be raised by _____________
(a) Chairman
(b) Secretary
(c) Member
Answer:
(c) Member

Question 6.
One share one vote’ is a voting method following under _____________
(a) voting by show of hands
(b) voting by ballot
(c) voting by-poll
Answer:
(c) voting by-poll

Question 7.
A motion which is changed due to an amendment is called _____________ motion.
(a) Formal
(b) Substantive
(c) Closure
Answer:
(b) Substantive

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Agenda	(1) Alteration in the original motion
(b) Amendment	(2) Extra vote for chairman
(c) Point of Order	(3) Rejected a motion
(d) Casting Vote	(4) Objection raised by a member in the meeting
(e) Resolution	(5) List of items to be transacted at the meeting
	(6) Alteration in a Resolution
	(7) Suggestion given by the member in the meeting
	(8) Accepted Motion
	(9) Only important items to be discussed at the meeting
	(10) Extra vote for Directors

Answer:

Group ‘A’	Group ‘B’
(a) Agenda	(5) List of items to be transacted at the meeting
(b) Amendment	(1) Alteration in the original motion
(c) Point of Order	(4) Objection raised by a member in the meeting
(d) Casting Vote	(2) Extra vote for chairman
(e) Resolution	(8) Accepted Motion

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

Question 1.
A list of items to be transacted at the meeting.
Answer:
Agenda

Question 2.
The minimum number of members to be present for a valid meeting.
Answer:
Quorum

Question 3.
The person who presides over the meeting.
Answer:
Chairman

Question 4.
A proposal is put before the meeting for discussion and decision.
Answer:
Motion

Question 5.
A motion finally passed at the meeting.
Answer:
Resolution

Question 6.
An alteration was proposed to the original motion.
Answer:
Amendment

Question 7.
A motion which brings to the notice of the Chairman the irregularities at the meeting.
Answer:
Point of Order

Question 8.
A person appointed by a member to attend and vote at the meeting.
Answer:
Proxy

Question 9.
An extra vote for the Chairman in case of tie-in votes.
Answer:
Casting vote

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

Question 1.
A proxy can vote under the poll method.
Answer:
True

Question 2.
Notice of the meeting must be given 14 days in advance.
Answer:
False

Question 3.
An agenda is sent along with the notice.
Answer:
True

Question 4.
Secretary has to sign the minutes book.
Answer:
False

Question 5.
Motion is a final decision of the meeting.
Answer:
False

Question 6.
Alteration can be done only by adding some new words.
Answer:
False

Question 7.
Point of Order is an objection raised by a member.
Answer:
True

Question 8.
Chairman can exercise casting votes in case of tie-in votes.
Answer:
True

Question 9.
A proxy cannot speak at the meeting.
Answer:
True

Question 10.
Secrecy can be maintained under voting by the show of hands method.
Answer:
False

Question 11.
A proxy need not be a member of the company.
Answer:
True

1E. Find the odd one.

Question 1.
Ordinary, Resolution, Special Resolution, Formal Motion.
Answer:
Formal Motion

Question 2.
Voting by voice, Voting by Ballot, Casting vote.
Answer:
Casting vote

Question 3.
Formal Motion, Substantive Motion, Resolution.
Answer:
Resolution

Question 4.
Voting by Poll, Voting by show of hands, Virtual voting.
Answer:
Virtual voting

Question 5.
Bare Statement Agenda, Draft Minutes Agenda, Minutes.
Answer:
Minutes

1F. Complete the sentences.

Question 1.
A person who presides over the meeting is known as _____________
Answer:
Chairman

Question 2.
Agenda is sent along with _____________
Answer:
Notice

Question 3.
Quorum for a private company is _____________
Answer:
2 members

Question 4.
A person responsible for proper conduct and to maintain order in the meeting is _____________
Answer:
Chairman

Question 5.
A motion which is changed due to amendment is called _____________
Answer:
Substantive motion

Question 6.
A resolution passed by simple majority is called _____________
Answer:
Ordinary Resolution

Question 7.
A special resolution, within 30 days of its passing must be filed with _____________
Answer:
Registrar of Companies

Question 8.
A representative of a member in a meeting is called _____________
Answer:
Proxy

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Voting by voice	……………………..
(2) …………………	21 clear days
(3) Up to 1000 shareholders	……………………
(4) ……………………..	15 days
(5) More than 5000 shareholders	…………………….
(6) ……………………	Writing
(7) Voting by Poll	…………………….
(8) ……………………	Objection
(9) Resolution by circulation	……………………
(10) …………………	Chairman

(Casting vote, Point of order, Directors, Motion, Secret method, Quorum is 5 members, Notice, Yes or No, Quorum is 30 members, Minutes)
Answer:

Group ‘A’	Group ‘B’
(1) Voting by voice	Yes or No
(2) Notice	21 clear days
(3) Up to 1000 shareholders	Quorum is 5 members
(4) Minutes	15 days
(5) More than 5000 shareholders	Quorum is 30 members
(6) Motion	Writing
(7) Voting by-poll	Secret method
(8) Point of order	Objection
(9) Resolution by circulation	Directors
(10) Casting vote	Chairman

1H. Answer in one sentence.

Question 1.
What is a quorum?
Answer:
Quorum is the minimum number of members required to be present for a valid meeting.

Question 2.
What is a proxy?
Answer:
A proxy is a person, who can attend and vote at the meeting on behalf of an absent member.

Question 3.
What is motion?
Answer:
A motion is a proposal put before the meeting for discussion and decision.

Question 4.
What is Resolution?
Answer:
A motion accepted in a meeting is called Resolution.

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

Question 1.
The minimum number of members required to be present at a meeting’s called proxy.
Answer:
The minimum number of members required to be present at a meeting’s called Quorum.

Question 2.
Director draft s the minutes of a meeting.
Answer:
Secretary draft s the minutes of a meeting.

Question 3.
Voting by show of hands is a capitalistic method.
Answer:
Voting by show of hands is a democratic method.

Question 4.
The ordinary resolution requires a 75% majority of votes.
Answer:
A special resolution requires a 75% majority of votes.

Question 5.
Agenda is a record of business transacted at the meeting.
Answer:
Minutes is a record of business transacted at the meeting.

Question 6.
A notice period of general meeting needs 7 clear days.
Answer:
A notice period of general meeting needs 21 clear days.

1J. Arrange in proper order.

Question 1.
(a) Chairman
(b) Agenda
(c) Amendment
Answer:
(a) Agenda
(b) Chairman
(c) Amendment

Question 2.
(a) Point of order
(b) Quorum
(c) Motion
Answer:
(a) Quorum
(b) Motion
(c) Point of order

2. Distinguish between the following.

Question 1.
Motion and Amendment
Answer:

Basis	Motion	Amendment
1. Meaning	Motion is a written proposal placed before the meeting for decision and discussion.	An amendment is an alteration or modification proposed to the original motion.
2. Purpose	The main purpose of the motion is to discuss and to make proper decisions.	The main purpose of an amendment is to revise or modify the main motion.
3. Right to speak	Every member can speak only once either in favour or against the motion.	Only those members who have not spoken on the original motion can speak on an amendment.
4. Effect	When a motion is accepted by the majority in the meeting, it becomes a resolution.	When an amendment is approved by the majority, it becomes a part of the resolution.
5. Mover	Mover of the motion can speak twice on his own motions.	Mover of the amendment can speak only once on his own amendment.

Question 2.
Ordinary Resolution and Special Resolution
Answer:

Basis	Ordinary Resolution	Special Resolution
1. Meaning	The resolution which is passed by a simple majority of votes is called Ordinary resolution.	The resolution which is passed by 3/4th majority of votes is called a Special resolution.
2. Majority	It requires more than 50% of the majority.	It requires at least a 75% majority of votes.
3. Filing	It is not filed with the Registrar of companies.	It is filed with the Registrar of companies within 30 days of passing the resolution.
4. Business transacted	It is necessary for routine and ordinary business.	It is necessary for special business.
5. Examples	Appointment of auditors, declaration of dividend, the election of directors, etc.	Change in the name of the company, alteration in object clause, reduction in spare capital, etc.

3. Answer in brief.

Question 1.
State the importance of the Agenda.
Answer:
Agenda is a list of items to be discussed or things to be done or the business to be transacted at the meeting. It is sent along with the notice. A secretary prepares the agenda in consultation with the Chairman.
Importance of Agenda:

• Members come to know about the matter to be discussed at the meeting.
• Members can come well prepared for the meeting, as the agenda is sent along with notice (21 clear days) before the meeting.
• The agenda helps the chairman to conduct the meeting efficiently.
• Items can be discussed as per the order mentioned in the agenda.
• All items are discussed at the meeting. Routine matters are discussed first, followed by special matters.

Question 2.
State the provisions regarding Quorum.
Answer:
Quorum is the minimum number of members required to be present for transacting a valid business. Without a quorum, the proceeding of the meeting becomes invalid. The quorum should be present throughout the meeting i.e. from beginning till the end of the meeting. Secretary must check the quorum before the commencement of the meeting.

Provision relating to Quorum (Companies Act, 2013)
(i) Quorum for General Meeting:

In the case of a Public Company: Quorum depends on a number of shareholders.

No. of shareholders	Quorum
Up to 1000	5 members
1000 – 5000	15 members
More than 5000	30 members

In the case of a Private Company: Two members personally present shall be the quorum for the meeting.

(ii) Quorum for Board Meeting:
According to the provision in the Companies Act, a quorum for the board meeting is l/3rd of its total directors or two directors whichever is higher. Any fraction should be rounded to one.

4. Justify the following statements.

Question 1.
Resolution cannot be amended
Answer:
(i) When a motion is accepted in a meeting it becomes a resolution.
(ii) When a motion is under discussion it can be amended.
(iii) Resolution is a final decision, it cannot be changed.
(iv) Motion is passed in the meeting after a required discussion on a particular matter.
(v) A resolution cannot be amended as it is a final decision taken in the meeting.
Thus, the resolution cannot be amended.

Question 2.
Quorum is required throughout the meeting.
Answer:

• a minimum number of members required for a valid meeting is a quorum.
• Members present in the beginning should be present during the entire period of the meeting.
• If the quorum is not present, the meeting stands adjourned for the next week same day, same time, and same place.
• In the absence of a quorum, if any meeting is conducted it is considered an invalid meeting.
• If any resolution is passed in the absence of a quorum, then such resolution is considered an invalid and illegal resolution.
• Thus, a quorum is required throughout the meeting.

Question 3.
The chairman has the right of casting vote.
Answer:

• A casting vote is a special type of vote given only by the Chairman of the meeting.
• It is an additional vote given to the Chairman in case of equality of votes in favour and against a motion.
• The chairman can use his vote only in the case of a tie.
• This means when the votes cast for and against the resolution are equal.
• It is a decisive vote as the final decision depends on the manner in which it is used by the Chairman.
• The purpose of the casting vote is to have some definite decision on the matter for discussion before the meeting.
• As per the provision in the Articles of Association, this vote can be exercised only by the chairman of the meeting.
• Hence, the chairman has the right of casting vote.
• Thus, the chairman has the right of casting vote.

Question 4.
Minutes of a meeting, once approved cannot be changed.
Answer:

• Minutes are the factual and official records of the proceedings of the meeting in the form of decisions and resolutions.
• Minutes are the legal evidence of the proceedings conducted in a meeting.
• They cannot be changed once they are finalized and confirmed by the Chairman.
• If any rectification is to be done and is necessary, it is carried out by passing a resolution in the next meeting and recording the same in the minute’s book.
• Thus, the minutes of a meeting, once approved cannot be changed.

Question 5.
The agenda is useful to the chairman of the meeting.
Answer:

• Agenda means a list of items to be done at the meeting.
• It is a programme of the meeting.
• It is prepared by the secretary in consultation with the chairman.
• It is a guideline to the chairman of the meeting.
• It enables the chairman to conduct the business of the meeting in an orderly manner.
• Routine matters which can be discussed and decided in fewer times are arranged first in the serial order, after that special business or time-consuming matters.
• This makes it easy for the chairman to go through it and discuss the matters.
• Thus, the agenda is useful to the chairman of the meeting.

5. Answer the following questions.

Question 1.
Explain the kind of motion.
Answer:
A motion is a proposal put before the meeting for discussion and decision. A person who proposes a motion is called a proposer or a mover of a motion.
Kinds of Motion:

• Formal Motion
• Substantive Motion

(i) Formal Motion:
Formal motions are moved for the purpose of preventing or delaying or speeding up discussion on a motion.
Types of Formal Motion:
(a) Closure Motion:
This motion is moved when sufficient time is spent on the discussion of a particular motion. Any member can propose that ‘The question be now put’ to vote. This is a closure motion. The main object of this motion is to avoid waste of time and to arrive at a quick decision. If this motion is put to vote and if the majority approves, no further discussion is permitted.

(b) Previous Question Motion:
The main purpose of this motion is to prevent discussion on the main motion. When a member feels that it is unwise to consider the main motion, they may move the previous question. The wording of this motion is ‘The question be not now put’. If it is carried, the discussion on the main motion is dropped. If the previous question is lost, the original motion is put to vote.

(c) Next Business Motion:
The purpose of the motion is similar to the previous question motion. The wording of the motion is ‘The meeting to proceed to next business. Such a motion is moved when a member feels that the main motion under discussion is of little importance and other items of importance remain to be transacted. If it is carried, the original motion is dropped at once and the meeting proceeds to the next business.

(d) Adjournment of Debate Motion:
The main aim of this motion is to postpone the debate. The wordings of this motion are ‘The debate on the subject is adjourned’. The mover of the motion feels that some extra information is needed for further discussion and the discussion should be delayed for some time. If this motion is carried, the debate will be adjourned. If it is lost, the debate continues.

(e) Adjournment of Meeting:
The main aim of this motion is to postpone the meeting for a particular period or indefinite period. The wording of this motion is ‘The meeting be now adjourned’. If it is carried the meeting is postponed to a future date. If it is lost, the meeting continues.

(ii) Substantive Motion:
A motion that is changed due to an amendment is called a substantive motion. When an amendment is passed, it is incorporated in the original motion and the substantive motion is put to vote. If it is passed, it becomes a resolution.

Question 2.
Explain the essentials of notice of a meeting.
Answer:
Notice is an advance intimation given by the company informing the day, date, time, and place of meeting and business to be transacted at the meeting. It is given in writing to all those who are entitled to receive it.
(i) Essentials of the notice of a meeting:

• Nature and type of meeting.
• The exact day, date, time, and place of the meeting.
• The agenda of the meeting.
• A statement of a member entitled to attend the meeting and appoint a proxy.
• Any intention to pass a special resolution must be mentioned in the notice.
• Statutory note and explanation, if any special business is to be transacted.

(ii) Proper authority to send notice:
The Board of Directors is the proper authority to send the notice. Under exceptional circumstances, members of the National Company Law Tribunal or Central Government may send a notice for the meeting.

(iii) Proper authority to receive notice:

• All shareholders in case of the shareholders meeting.
• Auditors, in case of Annual General Meeting.
• The legal representative of the deceased or insolvent person.
• First name in Register of members, in case of joint holders.
• All directors in case of Board Meeting and General Meeting.

(iv) Period of Notice:
In the case of a general meeting, notice must be sent 21 clear days before the actual meeting (day of sending and day of meeting are excluded). In the case of a Board meeting, 7 days’ notice is required.

(v) Modes of serving notice:

• It can be sent either personally or by ordinary post at the registered address of the member in India.
• If no registered address in India, then it can be sent to the address, given by the members.
• It can be sent by registered post if requested by the member. Charges for the same are to be paid in advance.
• It is to be advertised in all leading newspapers, (English and regional language) where the registered office is situated.
• It can also be sent through electronic mode i.e. email.

(vi) A statement to be sent along with notice:
In case any special business is to be transacted in the general meeting, an explanatory statement to that effect is to be annexed to the notice.

(vii) Omission to give notice:
If notice is not served to one or more members deliberately, the meeting stands invalid.

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

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

Question 1.
DIN means __________
(a) Director Identification Number
(b) Secretary Identification Number
(c) Doctor Identification Number
Answer:
(a) Director Identification Number

Question 2.
Private company requires __________ Directors.
(a) 3
(b) 2
(c) 1
Answer:
(b) 2

Question 3.
COO means __________
(a) Chief Operating Officer
(b) Chief Financial Officer
(c) Chief Organization Officer
Answer:
(a) Chief Operating Officer

Question 4.
CS means __________
(a) Company Secretary
(b) Company Standard
(c) Chief Store Manager
Answer:
(a) Company Secretary

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) DIN	(1) Whole-time Director
(b) Executive Director	(2) Chief Executive Officer
(c) Key Managerial Personnel	(3) Alternate Director
(d) Company Secretary	(4) Director Identification Number
(e) Chief Financial Officer	(5) Member of ICWA
	(6) Officer responsible for Company’s finance
	(7) Member of ICSI
	(8) Officer responsible for Company’s Management

Answer:

Group ‘A’	Group ‘B’
(a) DIN	(4) Director Identification Number
(b) Executive Director	(1) Whole-time Director
(c) Key Managerial Personnel	(2) Chief Executive Officer
(d) Company Secretary	(7) Member of ICSI
(e) Chief Financial Officer	(6) Officer responsible for Company’s finance

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

Question 1.
DIN means.
Answer:
Director Identification Number

Question 2.
This is an audit that checks the compliance of the company.
Answer:
Secretarial Audit

Question 3.
It aims to achieve and integrate corporate practices to all companies.
Answer:
Secretarial Standard

Question 4.
As a principal officer of the company.
Answer:
Secretary

Question 5.
A Director who is in Whole Time Employment of the company.
Answer:
Whole Time Director

Question 6.
A first secretary appointed by the promoter.
Answer:
Pro-tem Secretary

Question 7.
Written proceeding of the meeting.
Answer:
Minutes

Question 8.
A person holding shares of nominal value ₹ 20,000.
Answer:
Small shareholders

Question 9.
Shares are required to be bought by the directors.
Answer:
Qualification shares

Question 10.
The person nominated on the Board by a dominant shareholder.
Answer:
Nominee Director

Question 11.
The director is in full-time employment with the company.
Answer:
Executive Director or Whole Time Director

Question 12.
Director designated as ‘Outside Director’.
Answer:
Nonexecutive Director

Question 13.
Fees paid to Directors for attending a Board meeting
Answer:
Sitting fees

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

Question 1.
A Directors include whole-time employment of company is called WTD.
Answer:
True

Question 2.
DIN is required for Directorship.
Answer:
True

Question 3.
Secretarial audit checks the compliance of the company.
Answer:
True

Question 4.
Secretary is not the principal officer of the company.
Answer:
False

Question 5.
Not providing guidance to the Board of Directors of the company is a general duty of the secretary.
Answer:
False

1E. Find the odd one.

Question 1.
Directors as Agents, as managing partners, As a company secretary.
Answer:
As a company secretary

Question 2.
First Directors, Casual vacancy, Audit.
Answer:
Audit

Question 3.
Chief Executive Officer, Chief Financial Officer, Promoter.
Answer:
Promoter

1F. Complete the sentences.

Question 1.
Statutory duties of secretary is to make all meetings __________
Answer:
minutes

Question 2.
There is no prohibition on having more than one in a company __________
Answer:
Managing Director

Question 3.
CFO is __________
Answer:
Chief Financial Officer

Question 4.
WTD means __________
Answer:
Whole Time Director

Question 5.
KMP is __________
Answer:
Key Managerial Personnel

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Managing Director	…………………………..
(2) …………………….	ICSI passed
(3) Manager	…………………………..
(4) Secretarial Audit	Checks the Legislations

(Substantial powers, Fulltime employee, Secretary, Checks the legislations)
Answer:

Group ‘A’	Group ‘B’
(1) Managing Director	Substantial powers
(2) Secretary	ICSI passed
(3) Manager	Full-time employee
(4) Secretarial Audit	Checks the Legislations

Question 2.

Group ‘A’	Group ‘B’
(1) First Director	……………………….
(2) …………………….	Key Managerial Personnel
(3) Pro-tem Secretary	……………………….
(4) Secretarial Standards	……………………….

(First Secretary, Appointed by the Promoter, Formulated by ICSI, Whole Time Director)
Answer:

Group ‘A’	Group ‘B’
(1) First Director	Appointed by the Promoter
(2) Whole Time Director	Key Managerial Personnel
(3) Pro-tem Secretary	First Secretary
(4) Secretarial Standards	Formulated by ICSI

1H. Answer in one sentence.

Question 1.
What do you mean by Nominee Director?
Answer:
The person nominated on the board by a major/dominant shareholder is known as the nominee director.

Question 2.
What do you mean by Independent Director?
Answer:
Independent Director is a director other than managing director, whole-time director, or a nominee director.

Question 3.
Which Companies have to appoint a Woman Director?
Answer:
Every listed company and the public company has a paid-up share capital of Rs. 100 crores or more and whose turnover is Rs. 300 crores or more, have to appoint a Woman Director.

Question 4.
What do you mean by Alternate Director?
Answer:
It means a director who is nominated by the board to act in the place of a director in his absence.

Question 5.
Who is in full-time employment with the company?
Answer:
The executive director (WTD) is in full-time employment with the company.

Question 6.
What do you mean by Qualification shares?
Answer:
Qualification shares are those shares that are to be bought by the intending director, so as to become a director of a company.

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

Question 1.
RIN is the unique identification number of the director.
Answer:
DIN is the unique identification number of the director.

Question 2.
Whole-time Director is not involved in the day-to-day management of the company.
Answer:
Non-Executive Director is not involved in the day-to-day management of the company.

Question 3.
A company having paid-up share capital of ₹ 10 crores or more has to appoint a Woman Director.
Answer:
A company having paid-up share capital of ₹ 100 crores or more has to appoint a Woman Director.

Question 4.
Big Shareholder’s Directors is not a mandatory appointment.
Answer:
Small Shareholder’s Directors is not a mandatory appointment.

Question 5.
Directors are regarded as an employee of the company.
Answer:
Directors are regarded as an elected representatives of shareholders of the company.

Question 6.
A small shareholder is a person holding shares of a maximum of ₹ 50,000.
Answer:
A small shareholder is a person holding shares of a maximum of ₹ 20,000.

Question 7.
The manager needs to be the director of the company.
Answer:
The managing director needs to be the director of the company.

2. Explain the following Terms/Concepts.

Question 1.
Pro-tem Secretary
Answer:
The first secretary appointed by the promoters of the company is termed as Pro-tem Secretary. He may or may not be appointed as a regular secretary by the Board.

Question 2.
Sitting Fees
Answer:
Fees paid to the director for attending Board or Committee meeting is called as Sitting fees. It may extend up to Rs. one lakh also.

3. Answer in brief.

Question 1.
State the importance of DIN.
Answer:
Importance of DIN:

• It helps the investors to take accurate and appropriate decisions, as through DIN they get to know the composition of top management of the company.
• It helps to handle the problems arising due to the company creating fraud after raising capital from the investors.
• It helps to detect and handle offenses committed by a particular Director.

Question 2.
What are the general duties of a Company Secretary?
Answer:
General duties depend on the size and nature of the business.
Some of the general duties are as follows:

• To provide guidance to the Board and Director with respect to their duties, responsibilities, and powers.
• To represent before different regulators and authorities and discharge different duties under the Act.
• To assist the Board in the conduct of the affairs of the company.
• To assist and advise the Board in ensuring good corporate governance.
• To perform all the duties that may be assigned by the Board from time to time.

Question 3.
State the rights of a Company Secretary.
Answer:
Rights of a Company Secretary: Rights are given to the Secretary by the Companies Act, Board of Directors, and the Shareholders.
The rights of the Company Secretary are given below:

• Right to control and supervise the working of his department.
• Right to be indemnified by the company if any loss is suffered by Secretary while performing or discharging his duties.
• Right to sign a document requiring authentication.
• Right to get remuneration as an employee of the company.

Question 4.
State the importance of Secretarial Audit?
Answer:
Importance of Secretarial Audit:

• It provides a mechanism that monitors the compliance requirements.
• It detects errors and mistakes in compliance with companies’ rules and regulations mechanism.
• It prevents the company from the risk and losses due to non-compliance.
• It builds the confidence of regulators, management, and stakeholders.
• Investors feel relaxed that the company is following a disciplined approach towards management.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 1.
What are alcohols? How are they classified?
Answer:
Alcohols are the hydroxy derivatives of hydrocarbons in which one or more hydrogen atoms are replaced by hydroxyl group.

Examples : CH₃ – OH methyl alcohol, CH₃ – CH₂ – OH ethyl alcohol. Depending on the basis of hydroxyl groups present in a molecule, alcohols are classified into monohydric, dihydric, trihydric and polyhydric alcohols.

Question 2.
What are monohydric alcohols? How are they classified?
Answer:
Alcohols having only one hydroxyl group in their molecules are called monohydric alcohols. Monohydric alcohols are classified according to the type of hybridization of the carbon atom to which the hydroxyl group is attached.

(1) Alcohols containing C_sp³ – OH bond : In these alcohols -OH group is attached to a sp³ – hybridised carbon atom of alkyl group. These alcohols are represented as R-OH. They are further classified as primary, secondary and tertiary alcohols in which – OH group is attached to primary, secondary and tertiary carbon atoms respectively.

(a) Allylic alcohols : In these alcohols -OH group is attached to a sp³ -hybridised carbon atom next to the carbon-carbon double bond i.e., to allylic carbon. Allylic alcohols may be primary, secondary and tertiary alcohols.

(b) Benzylic alcohols : In these alcohols -OH group is attached to a sp³ -hybridised carbon atom next to an aromatic ring. Benzylic alcohols may be primary, secondary and tertiary alcohols.

(2) Alcohols containing C_sp³ – OH bond: In these alcohols -OH group is attached to a sp² – hybridised carbon atom,
i.e., vinylic carbon. These alcohols are also called vinylic alcohols.
e.g., CH₂ = CH – OH vinyl alcohol.

Question 3.
What are phenols (carbolic acids)?
Answer:
Hydroxy derivatives of aromatic hydrocarbons in which the hydroxyl group is directly attached to the aromatic ring are called phenols.
Examples:

Question 4.
What is phenol? OR Define carbolic acid.
Answer:
The hydroxy derivative of benzene in which the OH group is directly attached to benzene ring is called phenol.

Question 5.
How are phenols classified? Give suitable examples.
Answer:
Phenols are classified on the basis of number of hydroxyl (- OH) groups present in a molecule of phenol.
(1) Monohydric phenols : Phenols contain one hydroxyl group in their molecule.

(2) Dihydric Phenols : Phenols contain two hydroxyl groups in their molecule.

(3) Trihydric phenols : Phenols contain three hydroxyl groups in their molecule.

Question 6.
What are ethers? How are ethers classified?
Answer:
They are alkoxy derivatives of alkanes in which a hydrogen atom of alkane (R – H) is replaced by alkoxy group ( – O – R) and divalent oxygen atom is attached to two alkyl groups or two aryl groups or one alkyl and one aryl group. Ethers are organic oxides R – O – Ar, Ar – O – Ar.
E.g. R – O – R or C₂H₅ – O – C₂H₅.

Ethers are classified into two groups as follows :
(1) Simple or symmetrical ether : The ethers in which both alkyl (or aryl) groups attached to the oxygen atom are same are called simple ethers.
E.g. (R – O – R), CH₃ – O – CH₃, dimethyl ether; C₆H₅ – O -C₆H₅ diphenyl ether.

(2) Mixed or unsymmetrical ethers : The ethers in which the two alkyl (or aryl) groups attached to the oxygen atom are different are called mixed ethers.
E.g. (R – O – R), CH₃ – O – C₂H₅ ethyl methyl ether; C₂H₅ – O – C₆H₅ ethyl phenyl ether.

Question 7.
What is the general formula of ethers?
Answer:
The general formula of ethers is C_nH_{2n + 2} O. For example, dimethyl ether CH₃ – O – CH₃ has molecular formula C₂H₆O.

Common nomenclature system :
(1) In this system, monohydric alcohols (R — OH) are named as alkyl alcohols.
(2) According to the attachment of hydroxyl group to a carbon atom they are named with prefixes as n-(normal or primary) alcohol, sec-(secondary) alcohol, tert-(tertiary) alcohol.
(3) Alcohols with two hydroxyl groups are named as glycols.

Carbinol system : In this system alcohols are considered as derivatives of methyl alcohol which is called carbinol. The alkyl group attached to the carbon carrying – OH group are named in alphabetical order. Then the suffix carbinol is added.

For example :

IUPAC system of nomenclature :

• In this system, alcohols are named as alkanols.
• A longest continuous chain of carbon atoms containing – OH group is chosen as a parent hydrocarbon and alcohol is considered as a hydroxy derivative of this alkane.
• The carbon atoms are numbered from a terminal carbon atom nearest to a carbon atom attached to – OH group so that the position of – OH group is indicated by the lowest locant.
• ‘e’ of an alkane is preplaced by ‘oT, giving alkanol. The number of OH groups is indicated by prefix, di, tri, etc. before ‘oT. The positions of -OH groups are indicated by appropriate locants.
• The different substituents are arranged in the alphabetical order, and their positions are indicated by proper numbers.
• Their names are hyphened on either sides except the last substituent.
• For cyclic alcohols are named by using prefix cyclo to the parent alkane considering – OH group attached to carbon atom C – 1.

Question 8.
Give common and IUPAC names for the following :
Answer:

Question 9.
Classify the following alcohols as primary, secondary and tertiary and write their IUPAC names.
Answer:

Question 10.
Give the structures of the following :
(1) Ethanol
(2) Propan-l-ol
(3) 2-Methylbutan-l-ol
(4) Pentan-3-ol
(5) Butan-2-ol
(6) 3-Methylbutan-2-ol
(7) Hexan-l-ol
(8) 3-Methylpentan-3-ol
(9) 2-Methylpropan-2-ol
(10) Butan-l-ol (H-Butyl alcohol)
(11) 2-Methylpropan-l-ol
(12) 2,3-Dimethylbutan-l-ol
(13) 2, 3-Dimethylbutan-2-oI
(14) 2-MethyIhexan-l-ol
(15) 2,2,3-Trimethylpentan-3-ol
(16) 2,3,3-Trimethylbutan-2-ol
(17) 3-EthyI-4-methylpentan-l-ol.
Answer:

Question 11.
Give the IUPAC names of the following alcohols. Classify them as primary (1°), secondary (2°) and tertiary (3°) alcohols. Identify allylic and benzylic alcohols amongst them.
Answer:

Question 12.
Write all the possible structural isomers of alcohol having molecular formula C6H140. Give their IUPAC names. Classify them as primary, secondary and tertiary alcohols. Identify optically active alcohols amongst them.
Answer:

The optical isomers of C₆H₁₄O are (2), (3), (6), (7), (12).

Question 13.
Give the IUPAC names of the following compounds :
Answer:

Question 14.
Write IUPAC names of following alcohols :
Answer:

Question 15.
Write the structures of following alcohols : (1 mark each)
Answer:

The IUPAC system name of phenol is benzenol. The common name phenol is also accepted by IUPAC. The common names have prefixes ortho, meta and para in substituted phenols. IUPAC system uses the locant 2-, 3-, 4-, etc. to indicate the positions of substituents.

Question 16.
Write the IUPAC names of the following compounds :
Answer:

Question 17.
Write IUPAC name of the following compounds :
Answer:

Question 18.
Give the structures and IUPAC names of isomeric phenols represented by the molecular formula C₈H₁₀O.
Answer:

Common and IUPAC system of nomenclature of ethers

In the common system of nomenclature, the ethers are named by writing names of the alkyl groups attached to the oxygen atom in alphabetical order and word ether is added. If two alkyl groups are same, prefix di- is used. According too the IUPAC system of nomenclature, ethers are named as alkoxyalkanes. The larger alkyl group is considered to be parent alkane. The name of the smaller alkane is prefixed by the name of alkoxy group and its locant.

Question 19.
Give common name and JUPAC name for the I11owing ethers :
Answer:

Question 20.
Give the IUPAC name of the following ethers.
Answer:

Question 23.
Give the structures and IUPAC names of all metameric ethers represented by formula C₅H₁₂O.
Answer:

Question 24.
How many isomeric compounds can be represented by formula C₄H₁₀O?
Answer:
A compound with the molecular formula C₄H₁₀O can show two functional isomers as a monohydric alcohol and ether.

Isomers of C₄H₁₀O as alcohol :

Isomers of C₄H₁₀O as ether :

Hence, the total number of isomers of C₄H₁₀O are seven

Question 25.
Write the structural formula and IUPAC names of all possible isomers of the compound with molecular formula C₃H₈O.
Answer:
Possible isomers of C₃H₈O with structural formulae and IUPAC names :

Question 26.
Write structures of alcoholic and ether isomers of a compound having molecular formula C₇H₈O.
Answer:

Question 27.
How is alkyl halide converted into alcohol by using
(1) Aqueous NaOH (or KOH),
(2) Moist silver oxide?
Answer:
(1) When an alkyl halide (R – X), is boiled with aqueous NaOH (or KOH) an alcohol is obtained,

(2) Alkyl halide when heated with moist Ag₂O, undergoes hydrolysis and forms an alcohol.

Question 28.
How are following compounds prepared by hydrolysis of alkyl halides
(1) Ethanol
(2) Isopropyl alcohol
(3) tert-butyl alcohol
(4) methyl alcohol
(5) butan-2-ol?
Answer:
(1) Ethanol When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.

(2) Isopropyl alcohol : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.

(3) Tert-butyl alcohol : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.

(4) Methyl alcohol : When methyl bromide (bromomethane) is heated with aq KOH, it is hydrolysed to methyl alcohol (methanol).

(5) Butan-2-ol : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.

Question 29.
How are following compounds prepared from alkyl halides using moist silver oxide?
(1) Ethanol
(2) Propan-2-ol.
Answer: ‘
(1) Bromoethane (C₂H₅Br) when boiled with moist Ag₂O undergoes hydrolysis and forms C₂H₅OH.

(2) When 2-chloropropane is boiled with moist Ag20, propan-2-ol is formed.

Question 30.
What is hydration of alkenes or olefins? How is it carried out? Explain with an example.
OR
How are alcohols prepared from alkenes?
Answer:
The addition of a water molecule across the double bond in an alkene is called hydration of alkenes or olefins.

Hydration does not take place directly. It is carried out by passing an alkene through cold and concentrated H₂SO₄ which forms deliquescent solid, alkyl hydrogen sulphate, which when boiled with water forms an alcohol.

Question 31.
How are the following compounds obtained by hydration of alkenes :
(1) Ethyl alcohol
(2) Isopropyl alcohol
(3) Tert-butyl alcohol?
Answer:
(1) When ethene is passed through cold 98 % H₂SO₄, ethyl hydrogen sulphate is formed, which on heating with water gives ethanol.

(2) Propene with cold 80% H₂SO₄ gives isopropyl hydrogen sulphate which further on boiling with water gives isopropyl alcohol.

(3) 2-Methylpropene (isobutylene) directly reacts with 50 % H₂SO₄ giving tert-butyl hydrogen sulphate, which when heated with water gives tert-butyl alcohol.

Question 32.
Identify C in the following reaction.

Answer:

Question 33.
Explain hydroboration-oxidation of alkene.
Answer:
When diborane is treated with alkene in the presence of tetrahydrofuran (THF) solvent, an addition product trialkyl borane is formed. Trialkyl borane is then oxidised with alkaline peroxide forms primary alcohol.

The addition of diborane to the double bond takes place in such a way that the boron gots attached to the less substituted carbon. The overall reaction gives Anti-Markovnikov’s product from unsymmetrical alkenes.

Question 34.
How is propan-l-ol prepared using diborane?
Answer:
When diborane is treated with propene, in the presence of THF an addition product tripropyl borane is formed. Tripropyl borane is then oxidised to propan-l-ol using hydrogen peroxide in the presence of dil NaOH.

The addition of diborane to the double bond takes place in such a way that the boron gets attached to the less substituted carbon. The alcohol formed by the addition of water to the alkene in a way opposite to the Markovnikov’s rule.

Question 35.
How is ethanol prepared using diborane?
Answer:
When diborane is treated with ethene in the presence of THF an addition product triethylborane is formed.Triethylborane is then oxidised with hydrogen peroxide to form ethyl alcohol.

Question 36.
Predict the major product when 2-methylbut-2-ene is converted into an alcohol in each of the following methods :
(1) acid catalysed hydration
(2) hydroboration by BH₃ – THF complex.
Answer:
(1) Acid catalysed hydration :

(2) Hydroboration by BH₃ – THF complex :

Question 37.
How are alcohols prepared from aldehydes and ketones?
OR
How are the following compounds obtained using Ni as catalyst and at high temperature
(1) Ethanol
(2) Propan-2-ol?
Answer:
Aldehyde and ketones are carbonyl compounds containing a carbonyl group  C = O. The reduction of the carbonyl group gives an alcohol.

(1) Primary alcohols are prepared by the reduction of aldehydes.

Example : When acetaldehyde is reduced with hydrogen in the presence of nickel as catalyst and at high temperature, ethyl alcohol (ethanol) is obtained.

(2) Secondary alcohols are prepared by the reduction of ketones.

Example : When acetone is reduced with hydrogen in the presence of nickel as catalyst and at high temperature, isopropyl alcohol (propan-2-ol) is obtained.

Question 38.
How will you convert carboxylic acids and esters to primary alcohols? Explain with suitable examples.
Answer:
Carboxylic acids and esters are not easily reduced by catalytic hydrogenation or by NaBH₄. However, relatively more reactive and an expensive LiAlH₄ is used to convert carboxylic acids and esters to primary alcohols. When acetic acid is reduced in the presence of LiAlH₄ and followed by their acid hydrolysis, ethyl alcohol is obtained.

When ethyl acetate is reduced in the presence of LiA1H₄ and followed by their acid hydrolysis. n-propyl alcohol and ethyl alcohol is obtained.

[Since LiA1H₄ is an expensive reagent. commercially acids are reduced to alcohol by converting them to esters, followed by their reduction. (catalytic hydrogenation)

Question 39.
How is Crotoflyl alcohol obtained from crotonaldehyde?
Answer:
When crotonaldehyde is reduced in the presence of lithium aluminium hydride, the produc obtained is hydrolysed to give crotonyl alcohol. Here. LiA1H₄ does not reduce carbon-carbon double bond.

Question 40.
WrIte the structure of aldehyde that yields

Answer:
The structure of aldehyde:

Question 41.
How are the following compounds prepared using Grignard reagent
(1) Ethanol
(2) Propan-l-ol
(3) Propan-2-ol
(4) 2-Methyl propan-2-ol?|
Answer:
(1) Ethanol : Formaldehyde on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on further hydrolysis with dilute HCl forms ethanol.

(2) Propanol-l-ol : Formaldehyde on reaction with Grignard reagent, C₂H₅ – Mg – I in dry ether forms a complex which on further hydrolysis with dilute HC1 forms Propan-l-ol.

(3) Propan-2-ol : An acetaldehyde on reaction with Grignard reagent in dry ether forms a complex which on further hydrolysis with dilute acid HC1, forms propan-2-ol.

(4) 2-Methyl propan-2-ol : Acetone on reaction with Grignard reagent in dry ether forms a complex which on further hydrolysis with dilute acid HCl, forms a tertiary butyl alcohol.

Question 42.
Give a mechanism of following reaction.

Answer:
Grignand reagent reacts with aldehyde or ketone to form an adduct which on hydrolysis with dil. acid gives the corresponding alcohol.

In the first step, the nucleophilic addition of Grigard reagent to the carbonyl group resulting in the formation of an adduct, which on hydrolysis yields an alcohol.

Question 43.
Write the structure of carbonyl compounds that can be converted by reduction methods into following alcohols :


Answer:

Answer:

Question 44.
Using Grignard reagent, suggest synthesis of following alcohols from aldehydes or ketones. Wherever possible, suggest more than one combination.

Answer:

(b) Synthesis of propan- 1-01:

(c) Synthesis of butan-2-ol:

(d) Synthesis of 2-methylhexan-2-oI:

Question 45.
How will you obtain butan-2-ol from
(1) Propanal
(2) butan-2-one
(3) but-2-ene?
Answer:
(1) Propanal : When propanal is treated with methyl magnesium iodide in the presence of dry ether, a complex is formed, which on acid hydrolysis butan 2-ol is obtained.

(2) Butan-2-one: When butan-2-one is hydrogenated at 413 K in the presence of catalyst finely divided nickel butan-2-ol is obtained.

(3) But-2-ene : When but-2-ene is passed through cold concentrated sulphuric acid, isobutyl hydrogen sulphate is formed. Isobutyl hydrogen sulphate on heating with water gives butan-2-ol.

Question 46.
Write the structure of aldehyde, carboxylic acid and ester that yield the following alcohol.

Answer:

Question 47.
How will you prepare?
(1) 2-Ntethylbutan-1-oI from an alkene.
Answer:

(2) CycIoheyImethanoI from a Grignard reagent.
Answer:

(3) 1-Phenyl ethanol from acelaldehyde.
Answer:

Question 48.
How are following conversions brought about?
(1) Benzyl chloride to benzyl alcohol. (NCERT)
OR
How is benzyl alcohol prepared from benzyl chloride?
Answer:

(2) Benzyl alcohol to Benzoic acid.
Answer:

(3) 1-Ethyl cyclohexanol from cyclohexanone.
Answer:

Question 49.
How is phenol (carbolic acid) prepared from chlorobenzene (Dow’s process)? OR
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
Preparation of phenol from chlorobenzene (Dow’s process) : Chlorobenzene is fused with NaOH at about 623 K under a pressure of about 150 atmospheres (1.5 x 10⁷ Nm^-2), when sodium phenoxide is formed. Sodium phenoxide is acidified with dil.HCl to obtain phenol.

Question 51.
How is phenol (carbolic acid) prepared from benzene sulphonic acid?
Answer:
Preparation of phenol from benzene sulphonic acid : Benzene sulphonic acid is neutralized with the requisite quantity of soda ash (Na₂CO₃) or NaOH and the solution is evaporated to obtain sodium benzene sulphonate salt. Dry sodium benzene sulphonate is fused with an excess of caustic soda (NaOH) at about 573 K when sodium phenoxide is formed. The fused mass of sodium phenoxide on treatment with dilute HC1 gives phenol.

Question 52.
How is phenol (carbolic acid) prepared from aniline (diazotization)?
OR
How is carbolic acid prepared from amino benzene?
Answer:
Preparation of phenol from aniline (diazotization) : When aniline is treated with sodium nitrite and hydrochloric acid (NaNO₂ + HC1) at low temperature (0°C – 5°C), benzene diazonium chloride is formed. This reaction is called diazotization. An aqueous solution of benzene diazonium chloride on warming with water or dil. H2S04 gives phenol.

Question 54.
Describe the physical properties of alcohols and phenols.
Answer:
The properties of alcohols and phenols are mainly due to the hydroxyl group.

(1) Nature of intermolecular forces : Due to presence of – OH groups, alcohols and phenols are polar molecules. The polar – OH groups are held together by the strong intermolecular forces i.e. hydrogen bonding.

(2) Physical State : Lower alcohols are colourless, toxic liquids having characteristic alcoholic odour. Pure phenol is colourless, toxic, low melting solid having characteristic carbolic or phenolic odour.

(3) Boiling Points : The boiling points of alcohols and phenols increase with increase in their molecular mass.
Methyl alcohol – 65 °C
Phenol -182° C
n-Butyl alcohol-118 °C
o-nitrophenol-217 °C

(4) Solubility : Solubility of alcohols and phenols in water due to their ability to form intermolecular hydrogen bonding.

Question 55.
Arrange the following compounds in order of their increasing boiling points.
Butan-2-ol, ethanol, pentan-l-ol, butan -l-ol, propan-l-ol, methanol.
Answer:
Methanol, ethanol, propan:l-ol, butan-2-ol, butan-1 -ol pentan-l-ol.

Question 56.
Explain the following :
(1) Ethanol has higher boiling point than ethane.
Answer:
(1) The hydroxyl group in alcohols is highly polar. The H-atom has partial positive charge and the oxygen atom has partial -ve charge. The hydroxyl group in ethanol is extensively hydrogen bonded.

(2) Large number of ethanol molecules associated together by intermolecular hydrogen bonding. The energy required to separate the molecules by breaking hydrogen bond into vapour state is higher. This results in increasing the boiling point of ethanol.

(3) On the other hand, in ethane (alkane) there is no hydrogen bonding between the molecules. The molecules of ethane are held together by weak van der Waals forces of attraction. Hence, these molecules can be easily separated. Thus, ethanol has higher boiling point than that of ethane.

(2) Methanol is more soluble in water than propan-l-ol.
Answer:
(1) Methanol being lower members of alcohols is more soluble in water, but as the size of an alkyl group or molecular weight of alcohol increases the solubility decreases.

(2) The solubility of methanol in water is due to polar characters of alcohols (R – O^-δ -H^+δ ) and water (H^+δ – O^-δ – H^+δ).

(3) The solubility of methanol is due to the formation of intermolecular hydrogen bonding between polar molecules of methyl alcohol and water. Hence, methyl alcohol is a associated liquid.

(4) In methyl alcohol, size of methyl group being very small, -OH group constitutes major part of the molecule giving more solubility. As size of alkyl group increases, the non-polar character increases the solubility decreases. Hence, methanol is more soluble in water than propan-l-ol.

Question 57.
Which of the following pair is more acidic and why?

Answer:
Due to high electronegativity of sp²-hybridized carbon, electron density on oxygen in phenol (I) decreases. This increases the polarity of O – H bond and results in more ionization of phenol than that of cyclohexyl alcohol (II). Therefore, phenol is more acidic than cyclohexyl alcohol.

Answer:
In p-nitrophenol, nitro group (N02) is an electron-withdrawing group present at ortho position which enhances the acidic strength (- I effect). The O – H bond is under strain and release of proton (H⁺) becomes easily. Hence, o-nitrophenol is more acidic than phenol.

Question 58.
Draw intramolecular hydrogen bonding structures in the following compounds :
(a) o-nitrophenol
(b) o-hydroxy benzoic acid.
Answer:
(a) o-nitrophenol :

(b) o-hydroxy benzoic acid :

Question 59.
Explain laboratory test of alcohols and phenols.
Answer:
Laboratory test : Aqueous solution of alcohols and phenols can be tested with litmus paper. Aqueous solution of alcohols is neutral to litmus (neither blue nor red litmus change colour). Aqueous solutions of phenols turn blue litmus red. Thus, phenols have acidic character.

Question 62.
Write the action of aq NaOH on phenol and the product obtained is acidified.
Answer:
Phenols dissolve in aqueous NaOH by forming water soluble sodium phenoxide and are reprecipitated/ regenerated as phenols on acidification with HC1.

Question 63.
Explain the action of sodium on ethanol.
Answer:
When cthanol is treated with sodium metal, sodium ethoxide is formed and hydrogen gas is liberated.

Liberation of H₂ gas is used to detect the presence of alcoholic -OH group of a molecule.

Question 64.
How are the following compounds obtained from alcohols using HCI:
(1) C₂H₅CI
(2) Isopropyl chloride
(3) tert-butyl chloride?
Answer:
(1) Ethyl alcohol, C2H5OH in the presence of Lucas reagent (ZnCl₂ + HCI conc.) forms ethyl chloride.

(2) Isopropyl alcohol reacts with Lucas reagent forms isopropyl chloride.

(3) Tert-butyl alcohol reacts with Lucas reagent forms tert-butyl chloride

Question 65.
What is Lucas reagent? What are its uses?
Answer:
Lucas reagent is composed of a mixture of concentrated HCl and Lewis acid. anhydrous ZnCI₂.

It is used to prepare alkyl chlorides and distinguish between 10, 2° and 3° alcohols.

Question 66.
How can alcohols be distinguished with the help of Lucas reagent?
Answer:
Lucas reagent is a mixture of concentrated HCI and anhydrous ZnCl₂. It is used to distinguish between primary, secondary and tertiary alcohols.
Alcohol with Lucas reagent forms an alkyl chloride, R -Cl which is insoluble and gives cloudiness and forms separate layer.

The time required for cloudiness to appear is based on the type of the alcohol and its reactivity :

(1) Primary alcohol : It reacts with the Lucas reagent very slowly and on heating forms alkyl chloride. The cloudiness and separation of a layer takes place after a long time on heating.

(2) Secondary alcohol : It reacts with the Lucas reagent much faster to form alkyl chloride. The cloudiness and separation of layer takes place slowly.

(3) Tertiary alcohol : It reacts immediately with the Lucas reagent at room temperature to form alkyl chloride. The cloudiness and separation of layer takes place instantaneously.

Question 67.
What is esterificatlon? how Is an ester obtained from alcohol or phenol?
Answer:
(1) When an alcohol or phenol is heated with a carboxylic acid in the presence of conc.sulphuric acid an ester is obtained. The reaction is called esterification. This is reversible reaction and the formation of an ester is favoured
using excess of alcohol in the presence of conc. H₂SO₄.

(2) Alcohols and phenols react with acid anhydrides in presence of acid catalyst to form eser.

(3) The reaction of alcohol and phenols with acid chloride is carried out in the presence of pyridine (base), which neutralizes HCl.

Question 68.
Explain the action of the following on ethanol :
(1) Acetic acid
(2) Acetic anhydride
(3) Acetyl chloride.
Answer:
(1) Acetic acid : When ethanol is treated with acetic acid in the presence of cone, sulphuric acid, ethyl acetate (ester) is formed.

(2) Acetic anhydride : When ethanol is treated with acetic anhydride in the presence of cone, sulphuric acid, ethyl acetate (ester) is formed.

(3) Acetyl chloride : When ethanol is treated with acetyl chloride in the presence of pyridine, ethyl acetate (ester) is formed.

(Pyridine neutralises HC1 formed during reaction)

Question 69.
Explain the action of the following phenol :
(1) Acetic acid
(2) Acetic anhydride
(3) Acetyl chloride
Answer:
(1) Acetic acid : When phenol is treated with acetic acid in the presence of cone, sulphuric acid, (ester) is formed.

(2) Acetic anhydride : When phenol is treated with acetic anhydride in the presence of cone, sulphuric acid, (ester) is formed.

(3) Acetyl chloride : When phenol is treated with acetyl chloride in the presence of cone, sulphuric acid, (ester) is formed.

(Pyridine neutralizes HC1 formed during reaction.)

Question 70.
What is the action of acetic anhydride on salicyclic acid?
Answer:
When acetic anhydride is treated with salicyclic acid in presence of glacial acetic acid, acetyl salicyclic acid (aspirin) is obtained.

Aspirin is a common analgesic, antipyretic drug. Reactions involving breaking of C – O bond of alcohol :

Question 71.
How are the following compounds prepared by using HBr from corresponding alcohols :
(1) Ethyl bromide
(2) Isopropyl bromide
(3) Tert-butyl bromide (2-Methyl-propan-2-ol)?
Answer:
(1) When ethyl alcohol is heated with HBr, ethyl bromide is formed. (HBr is prepared in situ by adding NaBr to HCl or H₂SO₄)

(2) Isopropyl alcohol on heating with HBr forms isopropyl bromide.

(3) Tert-butyl alcohol on heating with HBr forms tert-butyl bromide.

Question 72.
Explain the action of hydroiodic acid on the following :
(1) Propan-2-ol
(2) 3-Methyl butan-2-ol.
Answer:
(1) Propan-2-ol : When propan-2-ol is heated with hydroiodic acid, 2-iodopropane is formed.

(2) 3-Methyl butan-2-ol : When 3-Methyl butan-2-ol is heated with hydroiodic acid, 2-Iodo-2-methyl butane is obtained. Here, secondary alcohol is converted into a tertiary alkyl halide.

Question 73.
Describe the action of PCl₃ on
(1) ethanol
(2) Propan-l-ol
(3) Propan-2-ol.
Answer:
(1) Ethanol : When ethanol is treated with PCl₃, ethyl chloride is obtained.

(2) Propan-l-ol : When propan-l-ol is treated with PCl₃, n-propyl chloride is obtained.

(3) Propan-2-ol : When Propan-2-ol is treated with PCl₃, 2-Chloropropane is obtained.

Question 74.
Describe the action of PCl₅ on
(1) ethanol
(2) propan-2-ol.
Answer:
(1) Ethanol : When ethanol is treated with PCl₅, ethyl chloride is obtained.

(2) Propan-2-ol : When propan-2-ol is treated with PCl₅, 2-chloropropane is obtained.

Question 75.
Describe the action of SOCl₂ on
(1) ethanol
(2) propan-l-ol.
Answer:
(1) Ethanol : When ethanol is treated with SOCl₂ in the presence of pyridine, ethyl chloride is obtained.

(2) Propan-l-ol : When propan-l-ol is treated with SOCl₂ in the presence of pyridine, w-propyl chloride is obtained.

Question 76.
Explain dehydration of alcohols.
OR
What is dehydration of alcohols? Give the chemical reactions showing dehydration of primary (1°), secondary (2°) and tertiary (3°) alcohols.
Answer:
Removal of water from an alcohol is called dehydration of alcohol. Alcohols having a /i-hydrogen is heated with dehydrating agents like concentrated H₂SO₄ (or H₃PO₄ or P₂O₅ or Al₂O₃). The ease of dehydration of alcohols is in the following order : tert-alcohol (3°) > secondary (2°) > primary (1°)
(1) Primary (1°) alcohol is dehydrated by heating it with 95% H₂SO₄ at 453 K.

(2) Secondary alcohol (2°) is dehydrated by heating with 60% H₂SO₄ at 373 K.

(3) A ternary alcohol can be easily dehydrated by heating with 20% H₂SO₄ at 363 K.

An alcohol can be dehydrated by passing vapours alcohols over heated alumina (Al₂O₃).

Question 77.
Explain oxidation of primary and secondary alcohols.
Answer:
(1) Primary alcohol on oxidation with CrO₃ forms aldehyde. However, a better reagent to bring about this oxidation is PCC (pyridinium chlorochromate).

(2) Secondary alcohol on oxidation with chromic anhydride (CrO₃) forms ketone.

Question 78.
What is the action of acidified K₂Cr₂O₇ on the following :

How are following alcohols distinguished :
(1) ethyl alcohol
(2) isopropyl alcohol
(3) tert-butyl alcohol?
OR
How will you distinguish primary, secondary and tertiary alcohols by oxidation process?
Answer:
Primary, secondary and tertiary alcohols are distinguished on the basis of their oxidation products, when their oxidation is carried out using K₂Cr₂O₇ and dil. H₂SO₄. Acidified potassium dichromate, K₂Cr₂O₇ is an oxidising agent.

(1) Ethyl alcohol is a primary alcohol and on oxidation, it first forms acetaldehyde, which on further oxidation forms acetic acid. In this both, aldehyde and acid have same number of carbon atoms.

(2) Isopropyl alcohol is a secondary alcohol and on oxidation it gives a ketone, acetone with the same number of carbon atoms. Acetone resists further oxidation as it involves breaking of C-C bond.

(3) The oxidation of tert-butyl alcohol is difficult, since it does not have a-hydrogen atom. It is oxidised by using acidic and stronger oxidising agents like KMn04, CrO₃ at high temperature, which dehydrate tertiary alcohol to alkene and then oxidise it to a ketone, acetone with less number of carbon atoms.

Question 79.
What happens when vapours of primary, secondary and tertiary alcohols are passed over heated copper at 573 K?
Answer:
When vapours of primary, secondary and tertiary alcohols are passed over heated copper at 573 K, dehydrogena¬tion of primary and secondary alcohol takes place while tertiary alcohols undergo dehydrogenation to given an alkene.

Primary alcohol :

Secondary alcohol :

Reactions of phenols :

Question 80.
(1) Explain the action of bromine in carbon disulphide (CS₂) on phenol (carbolic acid). OR Give equation of the reaction of bromine in CS₂ with phenol.
Answer:
When phenol is stirred at a low temperature with bromine dissolved in a polar solvent such as carbon disulphide or CCl₄ at (273 K), a mixture of o-bromophenol and p-bromophenol is formed, p-bromophenol is the major product.

Question 81.
Explain the action of bromine water on phenol (carbolic acid).
OR
Name the reagent used in the bromiriation of phenol to 2, 4, 6 tribromophenol.
Answer:
When phenol is treated with bromine water, a yellowish white precipitate of 2, 4, 6 – tribromophenol is formed.

Question 82.
Explain the action of dilute nitric acid on phenol (carbolic acid).
OR
Give equation of the reaction of dilute HNO₃ with phenol.
Answer:
When phenol is treated with dilute nitric acid, a mixture of o-nitrophenol and p-nitrophenol is formed. In this reaction, p-nitrophenol is formed as the major product.

Question 83.
Explain the action of cone, nitric acid (nitrating mixture) on phenol (carbolic acid).
OR
How is phenol converted into picric acid?
Answer:
When phenol is warmed with a mixture of cone, nitric acid and cone, sulphuric acid (a nitrating mixture or the mixed acid), 2, 4, 6 – trinitrophenol, commonly called picric acid, is formed.

Question 84.
Explain the action of concentrated sulphuric acid on phenol at different temperatures.
Answer:
(a) At room temperature : When phenol is treated with cone. H₂SO₄ at room temperature (about 300 K), o-phenol sulphonic acid is formed.

(b) At 373 K : When phenol is treated with cone. H₂SO₄ at about 373 K, p-phenol sulphonic acid is formed.

Question 85.
Explain (1) Kolbe’s reaction (2) Reimer-Tiemann reaction.
Answer:
(1) Kolbe’s reaction : When phenol reacts with sodium hydroxide, sodium phenoxide is obtained. Phenoxide ion being more reactive than phenol towards electrophilic substitution. Phenoxide undergoes electrophilic substitution with carbon dioxide at 398 K under pressure of 6 atm (a weak electrophile) forms salicylic acid as major product.

(2) Reimer-Tiemann reaction : Phenol is heated with chloroform along with aqueous NaOH, this is followed by acidification with dil. HC1 when salicyladehyde (2-hydroxy benzaldehyde) is formed as the major product, which can be separated from p-isomer by steam distillation. The stability of o-isomer is due to intramolecular hydrogen bonding.

Question 86.
Explain the action of zinc dust on phenol.
OR
How is phenol converted into benzene?
Answer:
When phenol is heated with zinc dust, benzene is obtained.

Question 87.
Explain the action of chromic acid on phenol.
OR
How is phenol converted into benzoquinone?
Answer:
When phenol is oxidised by chromic acid, a diketone, p-benzoquinone is formed. It is a conjugated diketone.

Question 88.
Explain catalytic hydrogenation of phenol.
Answer:
When a mixture of vapours of phenol and hydrogen is passed over nickel catalyst at 433 K., cyclohexanol is obtained.

Question 89.
How is diethyl ether (ethoxyethane) obtained from alcohol?
Answer:
When excess of ethyl alcohol is distilled with concentrated sulphuric acid (H₂SO₄) at 413 K, diethyl ether is formed.

Question 91.
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from Alkyl halides?
OR
How are simple ethers and mixed ethers prepared by Williamson’s synthesis?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained, this reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.

(A) Simple (Symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.

Sodium ethoxide on heating with ethyl bromide gives diethyl ether.

(B) Mixed (Unsymmetrical) ether : When an alkyl halide and sodium alkoxide or sodium phenoxide having different alkyl groups are heated, unsymmetrical ether (dialkyl ethers or alkyl aryl ether) is obtained.

Sodium ethoxide on heating with methyl bromide gives ethyl methyl ether.

Sodium phenoxide on heating with ethyl bromide gives ethyl phenyl ether.

Question 92.
How is anisole obtained from phenol?
OR
How is methoxy benzene prepared from carbolic acid?
Answer:
Phenol reacts with sodium hydroxide, sodium phenoxide is formed. When sodium phenoxide is heated with methyl iodide, anisole is obtained.

Question 93.
Explain the limitations for the preparation of unsymmetrical ethers.
OR
What care is to be taken in the preparation of unsymmetrical ethers by Williamson’s synthesis? Explain.
OR
Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.
Answer:
(1) In the preparation of unsymmetrical ethers by Williamson’s synthesis, the proper choice of the reactants namely alkyl halide and sodium alkoxide is necessary.

(2) The best yield of unsymmetrical ether is obtained when primary alkyl halide and tertiary alkoxide are heated, since primary alkyl halides are more susceptible to SN² reaction.

(3) Secondary or tertiary alkyl halide undergo a and fi (halogen and hydrogen) elimination reaction giving an alkene instead of an ether since a carbon atom is sterically hindered by bulky alkyl groups.

(4) For example : t-butyl methyl ether can be synthesised by reaction of methyl bromide with sodium t-butoxide.

Question 94.
State physical properties of ethers.
Answer:
(1) Dimethyl ether and ethyl methyl ether are gases. Other ethers are colourless liquids with pleasant odour.
(2) Lower ethers are highly volatile and highly inflammable substances.
(3) Boiling points of ethers show gradual increase with the increase in molecular mass.

(4) The solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass.

Question 95.
Explain, ethers posses a small net dipole moment.
Answer:
In ethers, bond angle is 110° and not 180°, bond dipole moments of the two C – O bonds do not cancel each other, therefore, ethers possess a smal net dipole moment, (for example, dipole moment of diethyl ether is 1.18 D)

Question 96.
Explain, the solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass.
Answer:
The solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass. This is because ethers can form hydrogen bonds with water through ethereal oxygen.

For example, diethyl ether and n-butyl alcohol have respective miscibilities of 7.5 and 9 g per 100 g of water.

Question 97.
Explain laboratory test for ethers.
Answer:
Ethers are neutral compounds in aqueous medium. Ethers do not react with bases, cold dilute acids, reducing agents, oxidizing agents and active metals. However, ethers dissolve in cold concentrated H₂SO₄ due to formation of oxonium salts.

This property distinguishes ethers from hydrocarbons.

Question 98.
What is the action of atmospheric oxygen on diethyl ether?
Answer:
When atmospheric oxygen combines with diethyl ether, peroxide of diethyl ether is obtained.

Question 99.
What is the action of dilute sulphuric acid on
(1) Dimethyl ether
(2) Diethyl ether
(3) Ethyl methyl ether?
(4) Anisole?
OR
Explain hydrolysis of ethers.
Answer:
Simple ethers on heating with dilute sulphuric acid under pressure undergoes hydrolysis to give alcohol.

(1) Dimethyl hydrolysis on hydrolysis give methanol.

(2) Diethyl ether on hydrolysis give ethanol.

(3) A mixed ether on heating with dii. H₂SO₄ under pressure undergoes hydrolysis to give mixture of two different alcohols.

Ethyl methyl ether on hydrolysis give a mixture of ethanol and methanol.

(4) Anisole on hydrolysis give a mixture of phenol and methanol.

Question 100.
State the combustion products of diethyl ether.
Answer:
The combustion products of diethyl ether are CO₂ and H₂O.

Question 101.
What is the action of phosphorus pentachloride on
(1) Diethyl ether
(2) Ethyl methyl ether
(3) Methyl phenyl ether (anisole)?
Answer:
(1) Diethyl ether : When diethyl ether is heated with ethyl methyl ether, ethyl chloride is formed.

(2) Ethyl methyl ether : When ethyl methyl ether is heated with phosphorus pentachloride, a mixture of ethyl chloride and methyl chloride is formed.

(3) Methyl phenyl ether (Anisole) : When methyl phenyl ether is heated with phosphorus pentachloride, a mixture of methyl chloride, chlorobenzene is formed.

Question 102.
Describe the action of hot concentrated HI on
(i) Dialkyl ether
(ii) Alkyl aryl ether.
Answer:
Ether reacts with excess of hot concentrated hydrogen halide to give two alkyl halide molecules.
R – O – R + HX → RX + R – OH
R – OH + HX → RX + H₂0
Alkyl aryl ether reacts with hot concentrated hydrogen halide to give phenol and alkyl halide.

Ethers with two different alkyl groups react with hot.conc. HI to give alkyl halides.
R – O – R’+ 2H – X → R – X + R’ – X + H₂O
The order of reactivity of HX is HI > HBr > HC1

Question 103.
What is the action of hot HI on isopropyl methyl ether?
Answer:
When isopropyl methyl ether is treated with excess of hot hydroiodic acid, a mixture of isopropyl iodide and methyl iodide is formed.

Question 104.
Describe the action of hydroiodic acid on the following :
(1) Diethyl ether.
Answer:
When diethyl ether (ethoxy ethane) is treated with hydroiodic acid, a mixture of ethanol and ethyl iodide is formed.

If excess of hydroiodic acid is available, then ethyl alcohol further reacts with hydroiodic acid at higher temperature to form ethyl iodide and water.

(2) Ethyl methyl ether.
Answer:
When ethyl methyl (methoxy ethane) is treated with hydroiodic acid, a mixture of ethyl alcohol and methyl iodide is formed.

If excess of hydroiodic acid is available, then ethyl alcohol further reacts with hydroiodic acid at higher temperature to form ethyl iodide and water.

(3) Methyl n-propyl ether.
Answer:
When methyl n-propyl ether (1-Methoxy propane) is treated with hydroiodic acid, a mixture of n-propyl alcohol and methyl iodide is formed.

If excess of hydroiodic acid is available then n-propyl alcohol further reacts with hydroiodic acid at higher temperature to form n-propyl iodide and water.

(4) Methyl phenyl ether (anisole).
Answer:
When methyl phenyl ether (anisole) is treated with hydroiodic acid, phenol and methyl iodide is formed. Here, phenol does not react further with HI because – OH group is attached to sp²-hybridised carbon atom and it cannot be replaced by iodide (nucleophile).

Question 105.
Draw the resonance structures of aromatic ethers.
Answer:
The alkoxy group in aromatic ether is a ring activating and ortho-, paradirecting group toward electrophilic aromatic substitution.

Resonance structures :

+ R Effect of – OR group results in increased electron density at the para- and two ortho-posotions (see resonance structures II, III and IV).

Question 106.
Describe the action of bromine in acetic acid on anisole.
OR
Write the equation of the reaction of bromination of anisole in ethanoic acid medium.
Answer:
When anisole is treated with bromine in acetic acid, /i-bromoanisole (major product) is obtained.

Question 107.
Describe the action of methyl chloride on anisole (Friedel-Crafts reaction).
OR
Write the equation of Friedel-Crafts reaction-alkylation of anisole.
Answer:
When anisole is treated with alkyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst, 4-Methoxy toluene is formed as major product. The alkyl groups are introduced at -ortho and -para positions in anisole, the reaction is known as Friedel-Crafts alkylation reaction.

Question 108.
Describe the action of Acetyl chloride on anisole (Friedel-Crafts acylation).
OR
Write the equation of the reaction Friedel-Crafts acylation of anisole.
OR
Write a note on Friedel-Crafts acylation.
Answer:
When anisole is treated with acetyl chloride in the presence of anhydrous aluminium chloride (a Lewis acid), 4-Methoxy acetophenone (major product) is obtained. The acetyl groups are introduced at -ortho and -para positions in anisole, the reaction is known as Friedel Craft’s acylation reaction.

Question 109.
Describe the action of cone. HNO₃ on anisole.
OR
Write the equation of nitration of anisole.
Answer:
When anisole is reacted with nitrating mixture (cone. HNO₃ + cone. Fl₂SO₄), a mixture of p-nitroanisole and o-nitroanisole is obtained.

Question 110.
An organic compound with the formula C₄H₁₀O₃ shows properties of ether and alcohol. When treated with an excess of HBr yields only one compound 1,2 dibromomethane. Write structural formula of ether and that of alcohol.
Answer:
When C₄H₁₀O₃ is treated with excess of HBr, a single compound 1,2-dibromoethane is formed.

Question 111.
An organic compound ‘A’ having molecular formula C₄H₁₀O does not react with sodium metal. On hydrolysis with dilute H₂SO₄ it gives only one organic compound ‘B’ The compound B on heating with red phosphorus and iodine gives compound ‘C’. The compound C can also be obtained from compound ‘A’ on heating with excess HI. Identify the compounds A, B and C.
Answer:
(1) The organic compound ‘A’ with molecular formula C₄H₁₀O may be an alcohol or ether.

(2) Since ‘A’ does not react with sodium (Na), it is not an alcohol. Hence ‘A’ may be an ether.
(3) Since an ether ‘A’ on hydrolysis with dilute H₂SO₄ gives only one compound ‘B’, the compound ‘A’ must be a symmetrical (simple) ether. Hence compound ‘A’ may be, C₂H₅ – O – C₂H₅

Question 112.
How will you affect the following two-step conversions? Diethyl ether into n-butane :
Answer:

Question 114.
Answer in one sentence/word.

(1) Name the alcohol that is used to make propan-2-one.
Answer:
The alcohol used to make propan-2-one is iso-propyl alcohol.

(2) Which is the first oxidation product of secondary alcohol?
Answer:
The first oxidation product of secondary alcohol is ketone.

(3) Name the alcohol that is used to make acetic acid.
Answer:
The alcohol that is used to make acetic acid is ethyl alcohol.

(4) Write the structure of cyclohexane-1, 4-diol.
Answer:

(9) Which of the following isomers is more volatile : o-nitrophenol or p-nitrophenol?
Answer:
The isomer o-nitrophenol with lower boiling point is more volatile.

(10) Identify the product of the following reaction.

Answer:

(11) Identify the product obtained by industrial synthesis of carboxylation of phenoxide ion followed by acidification.
Answer:
The product is phenol

(12) What is the product A obtained in the following reaction?

Answer:
The product is phenol

(13) Which positions are occupied by – NO₂ group during nitration of carbolic acid?
Answer:
– o – (ortho) and – p – (para) positions are occupied by – NO₂ group during nitration of carbolic acid.

(14) Write the name of reactants used for the preparation of ethyl-tert-butyl ether.
Answer:

are used for the preparation of ethyl-tert-butyl ether.

(15) The product formed in the reaction of reverse of dehydration of alcohol is
Answer:
The product formed in the reaction of reverse of dehydration of alcohol is alkene.

(16) Which rule is obeyed by hydroboration oxidation process?
Answer:
The rule obeyed by hydroboration oxidation process is opposite to the Markovnikov’s rule.

(17) Name the reagents for the complete hydroboration-oxidation reaction in step 1 and step 2.
Answer:
Step 1 : Diborane
Step 2 : Hydrogen peroxide and dil. NaOH.

(18) Write the name of the test by which methanol can be distinguished from ethanol.
Answer:
Iodoform test by which methanol can be distinguished from ethanol.

(19) Write the name of reactant used for preparation of phenol, which gives byproduct used as solvent.
Answer:
Reactant used in the preparation of phenol :

(20) Ether is a good solvent for Grignard reagent. Which property makes it a good solvent?
Answer:
Ether has a low polarity, this property makes it a good solvent.

(21) The C – O – C bond angle in dimethyl ether is
Answer:
The C – O – C bond angle in dimethyl ether is 110°.

Question 115.
State the uses of methyl alcohol.
Answer:
(1) Methyl alcohol is used as an industrial solvent for dissolving oils, fats, gums, etc.
(2) It is used for dry cleaning and preparation of perfumes and varnishes.
(3) It is used as antifreeze agent for automobile radiators at low temperature.
(4) It is used in the preparation of methyl chloride, dimethyl sulphate and formaldehyde.
(5) It is used to denature ethyl alcohol.

Question 116.
State the uses of ethyl alcohol. OR Write two uses of ethyl alcohol.
Answer:
(1) Ethyl alcohol is used as solvent for dyes, oils, perfumes, cosmetics and drugs.
(2) A mixture of 10-20% ethyl alcohol with petrol is used as motor fuel.
(3) A mixture of ethyl alcohol and calcium acetate in gel form is used as solid fuel.
(4) It is widely used in beverages.
(5) Since ethyl alcohol has low freezing point, it is used in thermometer.
(6) It is an effective tropical antiseptic therefore it is used in many mouth washes.
(7) It kills micro-organisms on wound surface and in the mouth but its low toxicity does not kill the cells of the skin or mouth tissues.
(8) It is used in the preparation of chloroform, iodoform, acetic acid and ethers.
(9) It is used as fuel.

Question 117.
Give the important uses of phenol.
OR
Write two uses of phenol.
Answer:
(1) Phenol is used in the preparation of phenol-formaldehyde polymer which is used in a plastic bakelite.
(2) It is used in the preparation of phenol-phthalein-an indicator and in certain dyes.
(3) It is used in the preparation of drugs such as salol, aspirin, etc.
(4) It is used in the preparation of dettol, which is an antiseptic.
(5) It is used in the preparation of 2,4-dichlorophenoxy acetic acid which is used as selective weed killer.
(6) It is used to prepare picric acid which is used as explosive.

Multiple Choice Questions

Question 118.
Select and write the most appropriate answer from the given alternatives for each sub-question:

1. Which one of the following is a tertiary alcohol?
(a) Pentan-l-ol
(b) Pentan-2-ol
(c) 2-Methylpentan-2-ol
(d) 3-Methylpentan-2-ol
Answer:
(c) 2-Methylpentan-2-ol

2. Which of the following is a primary alcohol?
(a) 3-ethyl-3-hexanol
(b) 2-butanol
(c) 3-methyl-l-butanol
(d) 1-hexanol
Answer:
(d) 1-hexanol

3. The molecular formula C₄H₁₀O represents
(a) aldehydes
(b) alcohols
(c) ethers
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

4. The general formula of primary, secondary and tertiary alcohols is
(a) C_nH_2nOH
(b) C_nH_2n-1OH
(c) C_nH_2n+1OH
(d) C_nH_n+1OH
Answer:
(c) C_nH_2n+1OH

5. Which of the following alcohols cannot be prepared by hydration of the corresponding alkene?
(a) Ethanol
(b) Propan- l-ol
(c) Propan-2-ol
(d) 2-Methylpropan-2-ol
Answer:
(b) Propan- l-ol

6. Which of the following compounds when treated with CH₃MgI in dry ether followed by the hydroly¬sis, will give Propan-2-ol?
(a) HCHO
(b) CH₃CHO
(c) CH₃CH₂OH
(d) CH₃COCH₃
Answer:
(b) CH₃CHO

7. To prepare 3-Ethylpentan-3-ol, the reagents needed are
(a) CH₃CH₂MgBr + CH₃COCH₂CH₃
(b) CH₃MgBr + CH₃CH₂CH₂COCH₂CH₃
(c) CH₃CH₂MgBr + CH₃CH₂COCH₂CH₃
(d) CH₃CH₂CH₂MgBr + CH₃COCH₂CH₃
Answer:
(c) CH₃CH₂MgBr + CH₃CH₂COCH₂CH₃

8. How is 1-propanol obtained?
(a) Using propanal
(b) Using propanone
(c) Using propene
(d) All of these
Answer:
(a) Using propanal

9.

Answer:
(a)

10. Ketone on reduction gives
(a) 1° alcohol
(b) 2° alcohol
(c) 3° alcohol
(d) all of these
Answer:
(b) 2° alcohol

11. Primary alcohols are prepared by catalytic hydro-genation of aldehydes in presence of ……………………………… as a catalyst
(a) aluminium bromide
(b) fluoroboric acid
(c) dry ether
(d) palladium
Answer:
(d) palladium

12. Lower member of alcohols are
(a) insoluble in water
(b) soluble in water
(c) insoluble in acetaldehyde
(d) insoluble in petrol
Answer:
(b) soluble in water

13. Which of the following compounds contain hydro¬gen bonds?
(a) Ethane
(b) Ethanol
(c) Methoxymethane
(d) Ethylene
Answer:
(b) Ethanol

14. Which of the following is a trihydric alcohol?
(a) n-propyl alcohol
(b) Glycerol
(c) Glycol
(d) Glycine
Answer:
(b) Glycerol

15. Aldehydes are first oxidation products of
(a) primary alcohols
(b) secondary alcohols
(c) tertiary alcohols
(d) carboxylic acids
Answer:
(a) primary alcohols

16. The oxidation product of alcohol depends on
(a) – OH group of an alcohol
(b) number of carbon atoms in alcohol
(c) number of hydrogen atoms attached to hydroxyl bearing carbon
(d) all of these
Answer:
(c) number of hydrogen atoms attached to hydroxyl bearing carbon

17. The order of ease of oxidation is
(a) primary > secondary > tertiary
(b) primary < secondary < tertiary
(c) primary > tertiary > secondary
(d) secondary > tertiary > primary
Answer:
(b) primary < secondary < tertiary

18. Ethyl alcohol \(\frac{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{443 \mathrm{~K}} \mathrm{~A} \stackrel{\mathrm{HBr}}{\longrightarrow}\) Ethyl bromide. Identify A.
(a) Ethyl hydrogen sulphate
(b) Ethylene
(c) Isopropyl hydrogen sulphate
(d) Acetic acid
Answer:
(b) Ethylene

19. One mole of PCl₅ reacts with one mole of ethyl alcohol to give
(a) 1 mole Cl₂
(b) 1/2 mole Cl₂
(c) 1 mole HCl
(d) 1/2 mole HCl
Answer:
(c) 1 mole HCl

20. When ethyl alcohol is reacted with sodium metal and the compound so formed is treated with ethyl bromide, the product formed is
(a) an acid
(b) an alkane
(c) an ether
(d) an ester
Answer:
(c) an ether

21. Dehydration occurs at the lowest temperature and concentration for
(a) methyl alcohol
(b) n-propyl alcohol
(c) iso-propyl alcohol
(d) tert-butyl alcohol
Answer:
(d) tert-butyl alcohol

22. One mole of sodium when reacts with one mole of methyl alcohol, gives
(a) one mole of oxygen
(b) one mole of hydrogen
(c) half mole of hydrogen
(d) half mole of oxygen
Answer:
(c) half mole of hydrogen

23. A compound ‘X’(C₃H₈O) on oxidation gives com-pound ‘ Y’(C₃H₆O₂), hence compound ‘X’ must be
(a) a ketone
(b) an aldehyde
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(c) a primary alcohol

24. The toxicity of alcohols
(a) increases with an increase in their molecular weight
(b) decreases with an increase in their molecular weight
(c) increases with a decrease in their molecular weight
(d) does not depend on their molecular weight
Answer:
(a) increases with an increase in their molecular weight

25. The structural formula of 2-phenyl ethanol is

Answer:
(c)

26. Compound that fails to give effervescence with NaHCO₃ is
(a) C₆H₅COO
(b) CH₃COOH
(c) C₆H₅OH
(d) Picric acid
Answer:
(c)

27. Phenol is a bifunctional compound because
(a) it gives reactions of hydroxyl group (-OH) as well as aromatic ring
(b) it is a strong acid
(c) it is insoluble in NaOH
(d) it is readily soluble in water
Answer:
(a) it gives reactions of hydroxyl group (-OH) as well as aromatic ring

28. Phenol gives characteristic colour with
(a) iodine solution
(b) bromine water
(c) ammonium hydroxide
(d) aqueous ferric chloride solution
Answer:
(d) aqueous ferric chloride solution

29. Ethanol and phenol are distinguished from each other by the action of
(a) neutral ferrous chloride
(b) neutral ferric chloride
(c) ferric hydroxide
(d) ferrous hydroxide
Answer:
(b) neutral ferric chloride

30. Which of the following compounds is used to prepare bakelite resin?
(a) Acetaldehyde
(b) Ethanol
(c) Phenol
(d) Methyl amine
Answer:
(d) Methyl amine

31. Increasing order of acid strength among p-methoxy-phenol, p-methyl phenol and p-nitro- phenol is as
(a) p-nitrophenol > p-methoxyphenol > p-methylphenol
(b) p-methylphenol > p-methoxyphenol > p-nitrophenol
(c) p-nitrophenol > p-methylphenol > p-methoxyphenol
(d) p-methoxyphenol > p-methylphenol > p-nitrophenol
Answer:
(d) p-methoxyphenol > p-methylphenol > p-nitrophenol

32. The IUPAC name of the compound is
(a) 3,3-dimethyl-l-hydroxy cyclohexane
(b) l,l-dimethyl-3-cyclohexanol
(c) 3,3-dimethyl-1-cyclohexanol
(d) l,l-dimethyl-3-hydroxy cyclohexane
Answer:
(c) 3,3-dimethyl-1-cyclohexanol

33. Which of the following is the most reactive towards electrophilic attack?

Answer:
(a)

34. In the following reaction, Ethanol \(\stackrel{\mathrm{PBr}_{3}}{\longrightarrow} \mathrm{A}\)\(\stackrel{\text { Alc-KOH }}{\longrightarrow} \text { B } \frac{\text { (i) } \mathrm{H}_{2} \mathrm{SO}_{4} \text {, room temp. }}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}, \text { heat }} \mathrm{C} \text {. }\). The product C is
(a) CH₃ – CH₂ – O – CH₂ – CH₃
(b) CH₃ – CH₂ – OSO₃H
(c) CH₃CH₂OH
(d) CH₂ = CH₂
Answer:
(c) CH₃CH₂OH

35. What is the general formula of ethers?
(a) C_nH_2nO
(b) C_nH_2n+2O
(c) C_nH_2n-1O
(d) C_nH_2n-2O
Answer:
(b) C_nH_2n+2O

36. The linkage is present in
(a) proteins
(b) ketones
(c) ethers
(d) aldehydes
Answer:
(c) ethers

37. Methoxy ethane is the functional isomer of
(a) CH₃CHOHCH₃
(b) CH₃CH₂CH₂OH
(c) CH₃ – O – CH₃
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

38. An oxygen atom in ether is
(a) sp²-hybridized
(b) sp³-hybridized
(c) sp-hybridized
(d) sp²-d-hybridized
Answer:
(b) sp³-hybridized

39. C₅H₁₂O represents
(a) only an alcohol
(b) only an aldehyde
(c) only an ether
(d) an alcohol and an ether
Answer:
(d) an alcohol and an ether

40. Ether molecules are
(a) tetrahedral
(b) angular
(c) pyramidal
(d) diagonal
Answer:
(b) angular

41. Which one of the following compounds is not isomeric with Ethoxypropane?
(a) 1-Methoxypropane
(b) 2-Methoxypropane
(c) 2-Methylpropane-2-ol
(d) 2-Methylbutan-2-ol
Answer:
(d) 2-Methylbutan-2-ol

42. The IUPAC name of C₂H₅ – O – CH₂ – CH(CH₃)₂ is
(a) 1-Ethoxy-1-butane
(b) 2-Ethoxy-2-butane
(c) l-Ethoxy-2-methylpropane
(d) 3-Ethoxy-2-methylpropane
Answer:
(c) l-Ethoxy-2-methylpropane

43. The IUPAC name of CH₃OC₆H₅ is
(a) methoxy phenyl ether
(b) phenoxy methane
(c) methoxy benzene
(d) methyl phenyl ether
Answer:
(c) methoxy benzene

44. In CH₃(CH₂)₃ – O -CH₃, the parent hydrocarbon of large alkyl group is
(a) n-butane
(b) butane
(c) pentane
(d) n-pentane
Answer:
(a) n-butane

45. Which one of the following alkyl halide gives best yield in Williamson’s synthesis?
(a) CH₃ – Br
(b) CH₃ – CH(Br) – CH₃
(c) CH₂ – CH(Br) – CH₃ – CH₃
(d) (CH₃)₃ C – Br
Answer:
(a) CH₃ – Br

46. Which one of the following ethers cannot be pre¬pared by using diazomethane?
(a) Dimethyl ether
(b) Diethyl ether
(c) Ethyl methyl ether
(d) t-Butyl methyl ether
Answer:
(b) Diethyl ether

47. The continuous etherification is carried out at
(a) 473 K
(b) 413 K
(c) 498 K
(d) 403 K
Answer:
(b) 413 K

48. Williamson’s synthesis is used for the preparation of
(a) only unsymmetrical ethers
(b) only symmetrical ethers
(c) both symmetrical and unsymmetrical ethers
(d) only methyl ethers
Answer:
(c) both symmetrical and unsymmetrical ethers

49. Which of the following ethers on hydrolysis gives two different products that are successive members of a homologous series?
(a) Methoxymethane
(b) Ethoxythane
(c) Methoxyethane
(d) 2-Methoxypropane
Answer:
(c) Methoxyethane

50. Which one of the following compounds dissolves in hot dilute sulphuric acid but does not react with sodium metal?
(a) Ethyl bromide
(b) Acetic acid
(c) Ethyl alcohol
(d) Diethyl ether
Answer:
(d) Diethyl ether

51. Identify ‘A’ in the following reaction :

Answer:
(c)

52. The C-O-C bond angle in an ether is
(a) 180
(b) 90°
(c) 104.5°
(d) 109.5°
Answer:
(d) 109.5°

53. Ether free from moisture and alcohol is known as ,
(a) dry ether
(b) absolute ether
(c) pure ether
(d) spirit ether
Answer:
(b) absolute ether

54. The geometry of ether is similar to
(a) ammonia
(b) methane
(c) water
(d) ethyne
Answer:
(c) water

55. Sodium metal does not react with
Answer:
C₂H₅ – O – C₂H₃

56. Ethers have boiling points
(a) lower than those of alkanes of comparable molecular masses
(b) higher than those of isomeric alcohols
(c) lower than those of isomeric alcohols
(d) higher than those of alkanes of comparable molecular masses
Answer:

57. Ether on hydrolysis gives
(a) aldehyde
(b) alcohol
(c) acid
(d) ester
Answer:
(c) acid

58. Which of the following reactions represent Williamson’s reaction?
(a) R – O – R’ + HI →
\(\text { (b) } \mathrm{R}^{\prime}-\mathrm{OH}+\stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\)

(d) R – O – Na + R’X →
Answer:
(c)

59. Which of the following reactions represent the continuous etherification process?

Answer:
(b)

60. Ethers on complete combustion produce
(a) an alcohol and water
(b) an alkene and water
(c) an alkane and an alkene
(d) carbon dioxide and water
Answer:
(d) carbon dioxide and water

61. Diethyl ether is used as a solvent in many organic reactions because it
(a) is liquid at room temperature
(b) has lower boiling point
(c) contains divalent oxygen atom
(d) is inert in nature
Answer:
(d) is inert in nature

62. Which of the following reagents can be prepared using ether as a solvent?
(a) Tollen’s reagent
(b) Grignard’s reagent
(c) Schiff’s reagent
(d) Millon’s reagent
Answer:
(b) Grignard’s reagent

63. Diethyl ether is used as
(a) a hypnotic
(b) an antiseptic
(c) an anaesthetic
(d) an antipyretic
Answer:
(c) an anaesthetic

64. To obtain fuel, diethyl ether is mixed with
(a) ester
(b) ethanal
(c) ethanol
(d) 2-propanone
Answer:
(c) ethanol

65. Which of the following alcohols is prepared by acid catalyzed hydration of alkenes?
(a) Butan-l-ol
(b) Propan-l-ol
(c) Ethanol
(d) Methanol
Answer:
(c) Ethanol

66. Which of the following alcohols can be prepared by direct hydration of corresponding alkene in presence of 50 % sulphuric acid?
(a) Butan-l-ol
(b) Butan-2-ol
(c) 2-Methylpropan-1 -ol
(d) 2-Methylpropan-2-ol
Answer:
(d) 2-Methylpropan-2-ol

67. Which of the following alcohols cannot be prepared by reduction of carbonyl compounds?
(a) Pentan-l-ol
(b) Pentan-2-ol
(c) 2-Methylpentan-2-ol
(d) 3-Methylpentan-2-ol
Answer:
(c) 2-Methylpentan-2-ol

68. Which of the following conversions explains the acidic nature of alcohols?

Answer:
(b)

69. Which of the following compounds gives 3-ethyl- pentan-3-ol by the action of ethyl magnesium iodide followed by acid hydrolysis?
(a) Propanone
(b) Butanone
(c) Pentan-2-one
(d) Pentan-3-one
Answer:
(d) Pentan-3-one

70. Benzyl phenyl ether reacts with hydrogen bromide to give
(a) benzyl bromide and phenol
(b) benzyl alcohol and bromobenzene
(c) benzyl bromide and bromobenzene
(d) benzyl alcohol and phenol
Answer:
(a) benzyl bromide and phenol

71. Ethers are considered as
(a) monoalkyl derivatives of water
(b) divalent oxygen atom is attached to two alkyl groups
(c) alkyl derivatives of fatty acids
(d) condensation products of acid and alcohol
Answer:
(b) divalent oxygen atom is attached to two alkyl groups

72. Which of the following compounds is not isomeric with ethoxyethane?
(a) 1-methoxypropane
(b) 2-methoxypropane
(c) 2-methylpropan-2-ol
(d) 2-methylbutan-2-ol
Answer:
(d) 2-methylbutan-2-ol

73. Which of the following compounds dissolves in hot dilute sulphuric acid but does not reacts with sodium metal?
(a) Ethyl bromide
(b) Acetic acid
(c) Ethyl alcohol
(d) Diethyl ether
Answer:
(d) Diethyl ether

74. Which of the following alcohol will have the fastest rate of dehydration?

Answer:
(c)

75. The phenol having lowest acidity is

Answer:
(b)

76. Which of the following reagents is best for the following conversion?

Answer:
(a)

77. 3-Methyl butane-2-ol on heating with HI gives
(a) 2 -iodo-3-methyl butane
(b) 2-iodo-2-methyl butane
(c) l-iodo-3-methyl butane
(d) l-iodo-2-methyl butane
Answer:
(b) 2-iodo-2-methyl butane

78. In phenol carbon atom attached to -OH group undergoes –
(a) sp³-hybridisation
(b) sp-hybridisation
(c) sp²-hybridisation
(d) No hybridisation
Answer:
(c) sp²-hybridisation

79. Which among the following reducing agents is ‘not’ used to reduce acetaldehyde to ethyl alcohol?
(a) Na-Hg and water
(b) Zn-Hg and cone. HCl
(c) H₂-Raney Ni
(d) Li-A1H₄/H⁺
Answer:
(b) Zn-Hg and cone. HCl

80. Identify the weakest acidic compound amongst the following :
(a) p-nitrophenol
(b) p-chlorophenol
(c) p-cresol
(d) p-aminophenol
Answer:
(d) p-aminophenol

81. Natalite is a mixture of
(a) diethyl ether and methanol
(b) diethyl ether and ethanol
(c) dimethyl ether and methanol
(d) dimethyl ether and ethanol
Answer:
(b) diethyl ether and ethanol

82. The alcohol used in thermometers is
(a) methanol
(b) ethanol
(c) propanol
(d) butanol
Answer:
(b) ethanol

83. Which of the following is the first oxidation product of secondary alcohol?
(a) Alkene
(b) Aldehyde
(c) Ketone
(d) Carboxylic acid
Answer:
(c) Ketone

84. When phenol is heated with cone. HNO₃ in presence of cone. H₂SO₄ it yields
(a) o-nitrophenol
(b) p-nitrophenol
(c) 2, 4, 6-trinitrophenol
(d) m-nitrophenol
Answer:
(c) 2, 4, 6-trinitrophenol