NCERT Exemplar Class 12 Chemistry Solutions Chapter 2 Solutions

NCERT exemplar class 12 Chemistry solutions chapter 2: MCQ (Type 1)

Question:1

Which of the following units is useful in relating the concentration of a solution to its vapour pressure?
(i) mole fraction
(ii) parts per million
(iii) mass percentage
(iv) molality
Answer:

The answer is option (I). Vapour pressure and the concentration (mole fraction) of solution are related. Raoult’s law states that the vapour pressure $(p_A)$ of a component is directly proportional to the mole fraction $(x_A)$ of the component.
$p_x\propto x_A$
where, $x_A=\frac{n_A}{n_A+n_B}$

Question:2

On dissolving sugar in water at room temperature solution feels cool to the touch. Under which of the following cases dissolution of sugar will be most rapid?
(i) Sugar crystals in cold water.
(ii) Sugar crystals in hot water.
(iii) Powdered sugar in cold water.
(iv) Powdered sugar in hot water.
Answer:

The answer is option (iv).

Dissolution is an endothermic process, which is why the solution is cool to the touch. Also, powdered sugars have a higher surface area than sugar crystals, which further promotes dissolution.

Question:3

At equilibrium, the rate of dissolution of a solid solute in a volatile liquid solvent is __________.
(i) less than the rate of crystallisation
(ii) greater than the rate of crystallisation
(iii) equal to the rate of crystallisation
(iv) zero
Answer:

The answer is option (iii). The rate of dissolution of a solid solute is equal to the rate of crystallisation at equilibrium.

Question:4

A beaker contains a solution of a substance ‘A’. Precipitation of substance ‘A’ takes place when a small amount of ‘A’ is added to the solution. The solution is _________.
(i) saturated
(ii) supersaturated
(iii) unsaturated
(iv) concentrated
Answer:

The answer is option (ii). Adding a small amount of solute in an already saturated solution will make the solution supersaturated and will lead to the precipitation of the solute.

Question:5

The maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon ____________.
(i) Temperature
(ii) Nature of solute
(iii) Pressure
(iv) Nature of solvent
Answer:

The answer is option (iii). Solids don’t compress significantly on application of pressure and thus have no impact to their solubility in solvent.

Question:6

Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________.
(i) low temperature
(ii) low atmospheric pressure
(iii) high atmospheric pressure
(iv) both low temperature and high atmospheric pressure
Answer:

The answer is option (ii). Due to lower atmospheric pressure at high altitude, oxygen concentration is lower in the blood and tissues of people living there.

Question:7

Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?
(i) Methanol and acetone.
(ii) Chloroform and acetone.
(iii) Nitric acid and water.
(iv) Phenol and aniline
Answer:

The answer is option (i). Methanol has hydrogen bonds. When acetone is mixed with it, some of the bonds break, weakening the methanol-methanol interaction. This is why they exhibit a positive deviation.

Question:8

Colligative properties depend on ____________.
(i) the nature of the solute particles dissolved in solution.
(ii) the number of solute particles in solution.
(iii) the physical properties of the solute particles dissolved in solution.
(iv) the nature of solvent particles
Answer:

The answer is option (ii). The number of solute particles determines its colligative properties.

Question:9

Which of the following aqueous solutions should have the highest boiling point?
(i) $1.0MNaOH$
(ii) $1.0MN{a}_{2}S{O}_4$
(iii) $1.0MN{H}_{4}N{O}_3$
(iv) $1.0MKN{O}_3$

Answer:

The answer is option (ii). As $1.0MN{a}_{2}S{O}_4$ furnishes the maximum number of ions, it should have the highest boiling point.

Question:10

The unit of ebulioscopic constant is _______________.
(i) $Kkgmo{l}^{-1}orK(molality{)}^{-1}$
(ii) $molkg{K}^{-1}or{K}^{-1}(molality)$
(iii) $kgmo{l}^{-1}{K}^{-1}or{K}^{-1}(molality{)}^{-1}$
(iv) $Kmolk{g}^{-1}orK(molality)$
Answer:

The answer is option (i)
$K_b=\frac{\mathrm{\Delta }{T}_b}{m}=\frac{K}{molk{g}^{-1}}$
The unit of ebullioscopic constant is $Kkgmo{l}^{-1}orK(molality{)}^{-1}$

Question:11

In comparison to a 0.01 M solution of glucose, the depression in the freezing point of a 0.01 M $MgC{l}_2$ solution is _____________.
(i) the same
(ii) about twice
(iii) about three times
(iv) about six times
Answer:

The answer is option (iii). Colligative properties only depend on the number of solutes. While a 0.01 M solution of glucose does not ionize, 0.01 M $MgC{l}_2$ solution ionizes to produce a single $M{g}^{2+}$ion and $2C{l}^{-}$ ions.

Question:12

An unripe mango was placed in a concentrated salt solution to prepare pickle shrivels because _____________.
(i) it gains water due to osmosis.
(ii) it loses water due to reverse osmosis.
(iii) it gains water due to reverse osmosis.
(iv) it loses water due to osmosis.
Answer:

The answer is option (iv). Due to osmosis, Water inside the mango (lower concentration) moves outside in a concentrated salt solution (higher concentration).

Question:13

At a given temperature, the osmotic pressure of a concentrated solution of a substance _____________.
(i) is higher than that at a dilute solution.
(ii) is lower than that of a dilute solution.
(iii) is the same as that of a dilute solution.
(iv) cannot be compared with the osmotic pressure of a dilute solution.
Answer:

The answer is option (i).
At a given temperature, the osmotic pressure of a concentrated solution of a substance is higher than that of a dilute solution.

Question:14

Which of the following statements is false?
(i) Two different solutions of sucrose of the same molality prepared in different solvents will have the same depression at the freezing point.
(ii) The osmotic pressure of a solution is given by the equation $\pi =CRT$ (where C is the molarity of the solution).
(iii) Decreasing order of osmotic pressure for 0.01 M aqueous solutions of barium chloride, potassium chloride, acetic acid and sucrose is $BaC{l}_2>KCl>C{H}_{3}COOH>sucrose.$
(iv) According to Raoult’s law, the vapour pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution.
Answer:

The answer is option (i). Colligative properties depend on the molarity of the solute and the solvent. Two different solvents (with the same molality of sucrose) will have different freezing points

Question:15

The values of Van’t Hoff factors for KCl, NaCl and $K_{2}S{O}_4$, respectively, are
_____________.
(i) 2, 2 and 2
(ii) 2, 2 and 3
(iii) 1, 1 and 2
(iv) 1, 1 and 1
Answer:

The answer is option (ii). While KCl and NaCl ionize to give 2 ions,$K_{2}S{O}_4$ gives 3 ions. So, van’t Hoff factors for KCl, NaCl and K2SO4 are 2,2 and 3, respectively

Question:16

Which of the following statements is false?
(i) Units of atmospheric pressure and osmotic pressure are the same.
(ii) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of lower concentration of solute to a region of higher concentration.
(iii) The value of the molal depression constant depends on the nature of the solvent.
(iv) Relative lowering of vapour pressure is a dimensionless quantity.
Answer:

The answer is option (ii). Solvent molecules move from a solution containing a high concentration of solute to one with a low concentration of solute.

Question:17

Value of Henry’s constant K_H ____________.
(i) increases with an increase in temperature.
(ii) decreases with an increase in temperature.
(iii) remains constant.
(iv) first increases, then decreases.
Answer:

The answer is option (i). There is an increase in the value of Henry’s constant with an increase in temperature.

Question:18

The value of Henry’s constant K_H is _____________.
(i) greater for gases with higher solubility.
(ii) greater for gases with lower solubility.
(iii) constant for all gases.
(iv) not related to the solubility of gases.
Answer:

The answer is option (ii). Gases having a lower solubility have a higher $K_H$.

Question:19

Consider the Figure and mark the correct option.
(i) Water will move from the side (A) to side (B) if a pressure lower than the osmotic pressure is applied on piston (B).
(ii) Water will move from the side (B) to the side (A) if a pressure greater than the osmotic pressure is applied on the piston (B).
(iii) Water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B).
(iv) Water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on the piston (A).

Answer:

The answer is option (ii). If the piston above B applies a pressure exceeding the osmotic pressure, water will move from the concentrated solution (side B) to the Fresh Water (side A).

Question:20

We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of Van’t Hoff factor for these solutions will be in the order______.
(i) i_A< i_B< i_C
(ii) i_A> i_B> i_C
(iii) i_A= i_B= i_C
(iv) i_A<i_B> i_C
Answer:

The answer is the option (iii). Due to complete dissociation of NaCl, the van’t Hoff factor will be the same for the three solutions.

Question:21

On the basis of the information given below, mark the correct option.
Information:
(A) Inthe bromoethane and chloroethane mixture, intermolecular interactions of A–A and B–B type are nearly the same as A–B type interactions.
(B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions.
(C) In a chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type interactions.
(i) Solution (B) and (C) will follow Raoult’s law.
(ii) Solution (A) will follow Raoult’s law.
(iii) Solution (B) will show a negative deviation from Raoult’s law.
(iv) Solution (C) will show a positive deviation from Raoult’s law
Answer:

The answer is the option (ii). In a Bromoethane (A) and Chloroethane (B) mixture, the interactions A-A, B-B and A-B are nearly equal, so the solution will be nearly ideal and thus follow Raoult’s law.

Question:22

Two beakers of capacity 500 mL were taken. One of these beakers, labelled as “A”, was filled with 400 mL water, whereas the beaker labelled “B” was filled with 400 mL of 2 M solution of NaCl. At the same temperature, both the beakers were placed in closed containers of the same material and the same capacity as shown in the Figure.
At a given temperature, which of the following statements is correct about the vapour pressure of pure water and that of NaCl solution?
(i) The vapour pressure in container (A) is more than that in container (B).
(ii) The vapour pressure in container (A) is less than that in container (B).
(iii) The vapour pressure is equal in both containers.
(iv) The vapour pressure in container (B) is twice the vapour pressure in container (A).

Answer:

The answer is the option (i). Vapour pressure of container (B) is lower as NaCl, being a non-volatile solute, reduces the vapour pressure.

Question:23

If two liquids A and B form a minimum boiling azeotrope at some specific composition then _______________.
(i) A–B interactions are stronger than those between A–A or B–B.
(ii) The vapour pressure of the solution increases because more molecules of liquids A and B can escape from the solution.
(iii) Vapour pressure of the solution decreases because less number of molecules of only one of the liquids escape from the solution.
(iv) A–B interactions are weaker than those between A–A or B–B.
Answer:

The answer is the option
(i) A–B interactions are stronger than those between A–A or B–B.

Question:24

4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is _____________.
(i) 0.004
(ii) 0.008
(iii) 0.012
(iv) 0.016
Answer:

The answer is the option (iv)

$M=\frac{n}{}V=0.02=\frac{n}{4}orn=0.08$
$M=\frac{n}{\text{(Mass of water in Kg)}}=\frac{0.08}{}5=0.016$

Question:25

On the basis of the information given below, mark the correct option.
Information: On adding acetone to methanol, some of the hydrogen bonds between methanol molecules break.
(i) At a specific composition, the methanol-acetone mixture will form a minimum boiling azeotrope and will show positive deviation from Raoult’s law.
(ii) At a specific composition, the methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.
(iii) At a specific composition, the methanol-acetone mixture will form a minimum boiling azeotrope and will show negative deviation from Raoult’s law.
(iv) At a specific composition, the methanol-acetone mixture will form a maximum boiling azeotrope and will show negative deviation from Raoult’s law.
Answer:

The answer is option (ii) At a specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult’s law.

Question:26

K_H value for Ar(g), CO₂(g), HCHO (g) and CH₄(g) are 40.39, 1.67, $1.83\times {10}^{-5}$ and 0.413 respectively.
Arrange these gases in the order of their increasing solubility.
(i) $HCHO<C{H}_4<C{O}_2<Ar$
(ii) $HCHO<C{O}_2<C{H}_4<Ar$
(iii) $Ar<C{O}_2<C{H}_4<HCHO$
(iv) $Ar<C{H}_4<C{O}_2<HCHO$
Answer:

The answer is option (iii). Solubility decreases with increasing value of KH. The increasing order of solubility of the gases is as follows: $Ar<C{O}_2<C{H}_4<HCHO$

NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: MCQ (Type 2)

Question:27

Which of the following factor(s) affect the solubility of a gaseous solute in the fixed volume of liquid solvent?
(a) nature of solute (b) temperature (c) pressure
(i) (a) and (c) at constant T
(ii) (a) and (b) at constant P
(iii) (b) and (c) only
(iv) (c) only

Answer:

The answer is the option (i, ii) solubility of a gaseous solute depends on the nature of the solute, pressure and temperature.

Question:28

Intermolecular forces between two benzene molecules are near to same strength as those between two toluene molecules. For a mixture of benzene and toluene, which of the following is not true?
(i) ${\mathrm{\Delta }}_{mix}H=zero$
(ii) ${\mathrm{\Delta }}_{mix}V=zero$
(iii) These will form a minimum boiling azeotrope.
(iv) These will not form the ideal solution.
Answer:

The answer is option (iii, iv) As interactions in a Benzene and Toluene solution are nearly identical, they will form an ideal solution.
${\mathrm{\Delta }}_{mix}H=0$, ${\mathrm{\Delta }}_{mix}V=0$

Question:29

Relative lowering of vapour pressure is a colligative property because _____________.
(i) It depends on the concentration of a non-electrolyte solute in a solution and does not depend on the nature of the solute molecules.
(ii) It depends on a number of particles of electrolyte solute in a solution and does not depend on the nature of the solute particles.
(iii) It depends on the concentration of a non-electrolyte solute in solution as well as on the nature of the solute molecules.
(iv) It depends on the concentration of an electrolyte or non-electrolyte solute in solution as well as on the nature of solute molecules.

Answer:

The answer is option (i, ii) Nature of solute particles is not a factor in determining the relative lowering of vapour pressure. It is only affected by the number of solute particles (non-electrolyte and electrolyte solutes)

Question:30

Van’t Hoff factor i is given by the expression _____________.
(i) i = Normal molar mass/Abnormal molar mass
(ii) i = Abnormal molar mass/ Normal molar mass
(iii) i= Observed colligative property/calculated colligative property
(iv) i = Calculated colligative property/ Observed colligative property
Answer:

The answer is option (i,iii)
$i=\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}ori=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$

Question:31

Isotonic solutions must have the same _____________.
(i) solute
(ii) density
(iii) elevation in boiling point
(iv) depression at the freezing point

Answer:

The answer is option (ii, iii)

Isotonic solutions must have the same
(ii) density
(iii) elevation in boiling point

Question:32

Which of the following binary mixtures will have the same composition in the liquid and vapour phases?
(i) Benzene – Toluene
(ii) Water-Nitric acid
(iii) Water-Ethanol
(iv) n-Hexane – n-Heptane

Answer:

The answer is option (ii, iii). Water-nitric acid and Water-Ethanol solutions have the same composition in liquid and vapour phases

Question:33

In isotonic solutions ________________.
(i) The solute and the solvent are both the same.
(ii) Osmotic pressure is the same.
(iii) Solute and solvent may or may not be the same.
(iv) A solute is always the same solvent may be different.
Answer:

The answer is option (ii, iii). Osmotic pressure is the same for isotonic solutions.

Question:34

For a binary ideal liquid solution, the variation in total vapour pressure versus the composition of the solution is given by which of the curves?

Answer:

The answer is option (I, iv) for the ideal solution.
$p_1\propto x_1$ and $p_2\propto y_2$

Question:35

Colligative properties are observed when _____________.
(i) A non-volatile solid is dissolved in a volatile liquid.
(ii) A non-volatile liquid is dissolved in another volatile liquid.
(iii) A gas is dissolved in a non-volatile liquid.
(iv) A volatile liquid is dissolved in another volatile liquid.
Answer:

The answer is option (i, ii) Colligative properties can be observed on the dissolution of a non-volatile solid or liquid in a volatile liquid

NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Short Answer Type

Question:36

Components of a binary mixture of two liquids A and B were being separated by distillation. After some time, the separation of components stopped and the composition of the vapour phase became the same as that of a liquid phase. Both components started coming in the distillate. Explain why this happened.
Answer:

As the composition of both the liquid and vapour phase is same and both components are coming in the distillate, this implies that they have formed an azeotropic mixture.

Question:37

Explain why on the addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases, while the addition of 1 mol of methyl alcohol to one litre of water decreases its boiling point.
Answer:

The boiling point of water increases on the addition of NaCl, as it is a non-volatile solute and, hence, reduces the vapour pressure of water and increases the boiling point. Unlike NaCl, Methyl alcohol is more volatile than water and thus increases the vapour pressure of water and reduces the boiling point.

Question:38

Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions.
Answer:

“Like dissolves like” means that a solute dissolves in a solvent if its intermolecular interactions are similar. In other words, we can say that polar solutes can dissolve in polar solvents, and non-polar solutes can dissolve in non-polar solvents.

Question:39

Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however, molarity is a function of temperature. Explain.
Answer:

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{Volume of solution in litres}}$
Since molarity is dependent on the volume of the solution, it changes with temperature. The other concentration terms, like mass percentage, ppm, mole fraction, and morality, involve moles or mass of components, which are independent of temperature.

Question:40

What is the significance of Henry’s Law constant K_H?
Answer:

According to Henry’s law:
$p=K_H\times x$
$K_H=\frac{p}{}x$
The solubility of a gas in a liquid decreases with increasing Henry’s law constant.

Question:41

Why are aquatic species more comfortable in cold water in comparison to warm water?
Answer:

For a given pressure, as the temperature decreases, the solubility of oxygen in water increases. This means that cold sea water has a higher concentration of oxygen, and aquatic species thrive more in cold water.

Question:42

(a) Explain the following phenomena with the help of Henry’s law.
(i) A painful condition known as bends.
(ii) Feeling of weakness and discomfort in breathing at high altitude.
(b) Why soda water bottle kept at room temperature fizz on opening?
Answer:

(a)

1. Deep-sea divers require compressed air for breathing underwater, which has both $N_2$ and $O_2$. In atmospheric conditions, $N_2$ isn’t soluble in blood, but at higher pressure (as you go deeper into the ocean, the pressure increases) it starts dissolving. When the diver comes back up, the pressure decreases and $N_2$ comes out of the body, leaving behind bubbles in the bloodstream, which restrict the flow of blood affecting nerve impulses. In worse cases, these bubbles can burst the capillaries and not let $O_2$ reach the tissues. This condition, known as bends, is very painful and can even be life-threatening.

2. In lower pressure (at higher altitudes), $O_2$ exhibits a lower partial pressure, resulting in lower oxygen concentration in blood and tissues. This condition is known as anoxia and leads to climbers becoming weak and losing the ability to think clearly.

(b) $C{O}_2$ isn’t very soluble in soft drinks at atmospheric pressure. So, to dissolve $C{O}_2$ in soft drinks, soda bottles are sealed under high pressure. However, when the bottle is opened, the pressure decreases suddenly and excess $C{O}_2$ fizzes out.

Question:43

Why is the vapour pressure of an aqueous solution of glucose lower than that of water?
Answer:

In pure liquid water, molecules of water cover the entire surface. When a non-volatile solute is dissolved in water, it replaces part of the surface water molecules. As a result, the number of solvent molecules which can escape also gets reduced, and thus, the vapour pressure of the solution reduces.

Question:44

How does sprinkling of salt help in clearing snow-covered roads in hilly areas? Explain the phenomenon involved in the process.
Answer:

Salt is spread over snow-covered roads to lower the freezing point enough to melt the ice (snow). This leads to the snow melting away and clearing the road.

Question:45

What is a “semi-permeable membrane”?

Answer:

Semipermeable membranes are membranes which only permit the flow of solvent molecules, but not the solute molecules. Only the solvent molecules move across the semipermeable membrane during osmosis and Reverse osmosis.

Question:46

Give an example of a material used for making a semipermeable membrane for carrying out reverse osmosis.
Answer:

For carrying out reverse osmosis, cellulose acetate, potassium ferrocyanide, etc. are used as semipermeable membranes.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Matching Type

Question:47

Match the items given in Column I and Column II.

Answer:

(i —> d), (ii —> c); (iii—> a); (iv —> b), (v —> f); (vi —> e)
i.Saturated solution: It is a solution where no more of the solute can be dissolved at the given temperature and pressure.
ii. Binary solution: It is a solution with two components.
iii. Isotonic solution: It is a solution having the same osmotic pressure at a given temperature as that of a given solution.
iv. Hypotonic solution: It is a solution whose osmotic pressure is less than that of a given solution.
v, Solid solution: It is a solution in a solid phase.
vi . Hypertonic solution: It is a solution whose osmotic pressure is greater than that of a given solution

Question:48

Match the items given in Column I with the type of solutions given in Column II.

Answer:

(i → e), (ii → c), (iii → d); (iv → b), (v → a)
i. Soda water: A solution of gas in $(C{O}_2)$ liquid(soft drink)
ii. Sugar solution: A solution of solid (sugar) in liquid (water)
iii. German silver: It is an alloy of Cu, Zn and Ni and a solid solution of solid in solid.
iv. Air: A solution of gas. Air is a mixture of various gases.
v. Hydrogen gas in palladium: Hydrogen gas in palladium is used as a reducing agent and is an example of a solution of gas in a solid.

Question:49

Match the laws given in Column I with expressions given in Column II.

Answer:

(i) → (c) (ii) → (e) (iii) → (d) (iv) → (a) (v) → (b)
i.Raoult’s law: $p=x_1{P}_1^0+x_2{P}_2^0$
ii. Henry’s law: $p=K_H.x$
iii. Elevation of the boiling point: $\mathrm{\Delta }{T}_b=K_{b}m$
iv. Depression in freezing point: $\mathrm{\Delta }{T}_f=K_{f}m$
v. Osmotic pressure: $\pi =CRT$

Question:50

Match the terms given in Column I with expressions given in Column II.

Answer:

(i → d), (ii → c), (iii →b), (iv → e), (v →a)

NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Assertion and Reason Type

Question:51

In the following questions, a statement of Assertion (A) followed by a statement of
Reason (R) is given. Choose the correct answer out of the following choices:
(a) Assertion and reason both are correct statements and reason is the correct explanation for Assertion.
(b) Assertion and reason both are correct statements but the reason is not the correct explanation for Assertion.
(c) Assertion is a correct statement but the reason is wrong.
(d) Assertion and reason both are incorrect.
(e) Assertion is wrong but reason is correct.
Assertion (A): The molarity of a solution in a liquid state changes with temperature.
Reason (R): The volume of a solution changes with a change in temperature.
Answer:

The answer is option (a). Molarity depends on volume, which changes with temperature. Hence, molarity also changes with a change in temperature.

Question:52

In the following questions, a statement of Assertion (A) followed by a statement of
Reason (R) is given. Choose the correct answer out of the following choices:
(a) Assertion and reason both are correct statements and reason is the correct explanation for Assertion.
(b) Assertion and reason both are correct statements but reason is not the correct explanation for Assertion.
(c) Assertion is a correct statement but reason is wrong.
(d) Assertion and reason both are incorrect.
(e) Assertion is wrong but reason is correct.
Assertion (A): When methyl alcohol is added to water, the boiling point of water increases.
Reason (R): When a volatile solute is added to a volatile solvent, an elevation in boiling point is observed.
Answer:

The answer is option (d). Addition of a volatile solute to a volatile solvent increases vapour pressure and decreases boiling point. Methyl alcohol and water interaction are an example of the same.

Question:53

In the following questions, a statement of Assertion (A) is followed by a statement of
Reason (R) is given. Choose the correct answer out of the following choices:
(a) Assertion and the reason both are correct statements and the reason is the correct explanation for the Assertion.
(b) Assertion and reason both are correct statements but reason is not the correct explanation for the Assertion.
(c) Assertion is a correct statement but reason is wrong.
(d) Assertion and reason both are incorrect.
(e) Assertion is wrong but reason is correct.
Assertion (A): When NaCl is added to water, a depression in the freezing point is observed.
Reason (R): The lowering of the vapour pressure of a solution causes depression in the freezing point.
Answer:

The answer is option (a). Addition of a non-volatile solute to a volatile solvent decreases vapour pressure and increases boiling point.

Question:54

In the following questions, a statement of Assertion (A) is followed by a statement of
Reason (R) is given. Choose the correct answer out of the following choices:
(a) Assertion and the reason are correct statements, and the reason is the correct explanation forthe Assertion.
(b) Assertion and reason both are correct statements, but the reason is not the correct explanation for the Assertion.
(c) Assertion is a correct statement, but the reason is a wrong statement.
(d) Assertion and reason both are incorrect statements.
(e) Assertion is a wrong statement, but the reason is a correct statement.
Assertion (A): When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from the pure solvent side to the solution side.
Reason (R): Diffusion of solvent occurs from a region of high-concentration solution to a region of low-concentration solution
Answer:

The answer is option (b). Assertion and reason both are correct statements, but the reason is not the correct explanation for the Assertion.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 2: Long Answer Type

Question:55

Define the following modes of expressing the concentration of a solution. Which of these modes are independent of temperature and why?
(i) w/w (mass percentage)
(ii) x (mole fraction)
(iii) V/V (volume percentage)
(iv) M (Molarity)
(v) w/V (mass by volume percentage)
(vi) m (Molality)
(vii) ppm (parts per million)
Answer:

1. $\text{w/w(Mass percentage)}=\frac{\text{Mass of component in the solution}}{\text{Total mass of the solution}}\times 100$

2. $\text{V/V(Volume percentage)}=\frac{\text{Volume of component in the solution}}{\text{Total volume of the solution}}\times 100$

3. $\text{w/V (mass by Volume percentage)}=\frac{\text{Mass of the solute}}{\text{Total volume of the solution}}\times 100$

4. $\text{ppm (parts per million)}=\frac{\text{Number of parts of component}}{\text{Total number of parts of all components}}\times {10}^6$

5. $\text{x( Mole fraction)}=\frac{\text{Number of moles of component}}{\text{Total number of moles of all components}}$

6. $\text{M (Molarity)}=\frac{\text{Moles of solute}}{\text{Volume of the solution in litre}}$

7. $\text{m (Molality)}=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Mass percentage, ppm, mole fraction and molality are only dependent on mass or number of moles making them independent of temperature changes.

Question:56

Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
(a) $CHC{l}_3(l)\text{and}C{H}_{2}C{l}_2(l)$
(b) $NaCl(s)\text{and}{H}_{2}O(l)$
Answer:

$p=p_A^o{x}_A+p_B^o{x}_B$
Where, $p_A^o,p_B^o$ are the vapour pressure of pure components (A) and (B)
And $x_A\text{and }{x}_B$ are the mole fractions of the components in the solution.

(b) On dissolving in water, the vapour pressure lowers. The vapour pressure of the solution is:
$p=p^o{x}_A$
Where $x_A$=mole fraction of solvent
$p^0$=vapour pressure of pure solvent
p=vapour pressure of the solution
Similarly,
$\frac{\mathrm{\Delta }p}{p^o}=x_B$
$\mathrm{\Delta }p=p^o-p$
$x_B$=mole fraction of solute

Question:57

Explain the terms ideal and non-ideal solutions in light of forces of interactions operating between molecules in liquid solutions.
Answer:

Ideal solutions: They obey Raoult’s law irrespective of the concentration. For an ideal solution,
${\mathrm{\Delta }}_{mix}H=0,{\mathrm{\Delta }}_{mix}V=0$
$\text{A-B interactions}\approx \text{A-A interactions and B-B interactions}$
Non-ideal solutions: They don’t follow Raoult’s law over the entire range of concentration
Positive deviations: Vapour pressure is higher than the calculated values
${\mathrm{\Delta }}_{mix}H=+ve,{\mathrm{\Delta }}_{mix}V=+ve$
A-B interactions < A-A interactions and B-B interactions
Negative deviations: Vapour pressure is lower than the calculated values
${\mathrm{\Delta }}_{mix}H=-ve,{\mathrm{\Delta }}_{mix}V=-ve$
A-B interactions>A-A interactions and B-B interactions

Question:58

Why is it not possible to obtain pure ethanol by fractional distillation? What general name is given to binary mixtures which show deviation from Raoult’s law and whose components cannot be separated by fractional distillation? How many types of such mixtures are there?
Answer:

Azeotropes are solutions which have the same composition of components in both the liquid and vapour phases and boil at a constant temperature. The components cannot be separated by fractional distillation as they have the same boiling point. There are two types of azeotropes

1. Minimum boiling azeotropes: They show a large positive deviation from Raoult’s law (A-B interactions are weaker than A-A and B-B interactions), e.g., an ethanol-water mixture

2. Maximum boiling azeotropes: They show large negative deviation from Raoult’s law (A-B interactions are stronger than A-A and B-B interactions), e.g., the solution having composition 68% HNO₃ and 32% water by mass.

Question:59

When kept in water, the raisin swells in size. Name and explain the phenomenon involved with the help of a diagram. Give three applications of the phenomenon.
Answer:

Raisins swell in water due to osmosis. Water moves from a place of lower concentration (Water holder) to a higher concentration (Raisin) through the skin of raisin, which acts as a semipermeable membrane.

Applications of the phenomenon

1. Water moves from the soil to the plant roots partly due to osmosis.

2. Preservation of meat against bacterial action by adding salt.

3. Preservation of fruits against bacterial action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis, shrivel, and die.

4. Reverse osmosis is used for the desalination of water.

Question:60

Discuss the biological and industrial importance of osmosis.
Answer:

Some of the biological and industrial importance of osmosis are as follows-

1. Osmosis is responsible for the circulation of water to all the body parts of animals

2. Water moves from soil to plant roots partly due to osmosis. The concentration of cell sap inside the root hair cells is higher compared to that of water present in the soil.

3. Water circulation inside the plant body from root to treetop is also because of osmosis.

4. Osmosis helps in the growth of plants and the germination of seeds.

5. Due to endosmosis, Red Blood Cells burst when they are placed in water.

6. Osmosis controls various functions of plants, e.g., the stretching of leaves and flowers, the opening, and the closing of flowers.

7. Salt and sugar in pickles prevent the growth of bacteria and fungi by osmosis and thus act as preservatives.

8. Endosmosis is responsible for the swelling of dead bodies under water.

9. Dried fruits and vegetables swell and return to their original form when placed in water. It is also due to the osmosis of water.

10. Edema: Tissues become puffy when a person ingests an excess amount of salt.

Question:61

How can you remove the hard calcium carbonate layer of the egg without damaging its semipermeable membrane? Can this egg be inserted into a bottle with a narrow neck without distorting its shape? Explain the process involved.
Answer:

This can be achieved as under:

1. Take a mineral acid solution and put an egg inside it, and leave it for 2 hours. Most of the outer shell will dissolve. Remove any remaining parts with your fingers.

2. Take a saturated solution (hypertonic) and place the egg in it for 3 hours. The egg’s size reduces as the egg shrivels due to osmosis.

3. Place the egg in a bottle with a narrow neck and fill water (hypotonic) it. The egg will regain its shape.

Question:62

Why is the molar mass determined by measuring a colligative property in case some solutes are abnormal? Discuss it with the help of the Van’t Hoff factor.
Answer:

Abnormal molecular masses are shown by the compounds which dissociate/associate in the solvent.

1. Association: Colligative properties depend on the number of particles in a solution. Certain compounds like benzoic acid or ethanoic acid dimerise in benzene due to hydrogen bonding, resulting in the reduction of the number of particles and thus, solutes show lower colligative properties.

2. Dissociation: Similarly, certain compounds like electrolytes (NaCl, KCl), etc dissociate into ions, increasing the number of particles, and thus, show a higher value of colligative property.

Van’t Hoff introduced a factor to account for association or dissociation, known as the Van’t Hoff factor.
$i=\frac{\text{Expected molar mass}}{\text{Abnormal molar mass}}$
$=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$
$=\frac{\text{Total number of moles of particles after association or disaasociation}}{\text{Total number of moles of particles before association or disaccioation}}$