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**Q.1. At 35°C, the vapor pressure of CS _{2} is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS_{2} in acetone has a total vapor pressure of 600 mm Hg. The false statement amongst the following is (2020)**

Since, in the given mixture, the solvent – solvent and solute – solute interaction is more than that of solvent – solute interaction, thus, the above mixture shows positive deviation from Raoult’s law.

Hence, statement (b) is incorrect.

(a) The volume of the solution increases and the volume of the solvent decreases.

(b) The volume of the solution decreases and the volume of the solvent increases.

(c) The volume of the solution and the solvent does not change.

(d) The volume of the solution does not change and the volume of the solvent decreases.

Due to the non-volatile solute in the mixture, there is a lowering of vapor pressure, thus, the volume of solution increases. Since, volume of container is constant, so on increasing the volume of solution, the volume of solvent alone decreases.

(I) X has higher intermolecular interactions compared to Y.

(II) X has lower intermolecular interactions compared to Y.

(III) Z has lower intermolecular interactions compared to Y.

The correct inference(s) is/are

(a) (I) and (III)

(b) (I)

(c) (II)

(d) (III)

Since, the boiling point of X is lower than Y, thus, the intermolecular interaction between molecules of X is less than that of Y.

Given, ΔT

We know

ΔT

Substituting the values in Eq. (1), we get

Thus, 1.75 g of NaCl should be added to 600 g of water to decrease the freezing point of water to –0.2°C

**Ans. **(2.18)**Solution.**

We know

ΔT_{f}* = *K_{f}m

⇒ 273 - T_{f} = 2 x 0.5

⇒ T_{f} = 272 K

Using Ideal gas equation,

pV = nRT (1)

Substituting the values in Eq. (1), we get

Since, ideal gas is in thermal equilibrium with aqueous solution of ethylene glycol

⇒ P_{1}V_{1} = P_{2}V_{2}**Q.6. Which one of the following statements regarding Henry's law is not correct? (2019)(a) Higher the value of K _{H} at a given pressure, higher is the solubility of the gas in liquids.(b) Different gases have different K_{H} (Henry's law constant) values at the same temperature.(c) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.(d) The value of K_{H} increases with increase of temperature and K_{H} is function of the nature of the gas.Ans.** (a)

The solubility of the gas in liquids decreases with the increase in value of K

(a) 48

(b) 32

(c) 64

(d) 16

Ans.

As we know,

ΔT

Total amount of water = 250 g

∴ The amount of water separated as ice = 250 - 186 = 64 g**Q.8. Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are 7 x 10 ^{3} Pa and 12 x 10^{3} Pa, respectively. The composition of the vapour is in equilibrium with a solution containing 40 mole percent of A at this temperature is: (2019)** (b)

(a) x_{A} = 0.37 ; x_{B} = 0.63

(b) x_{A} = 0.28 ; x_{B} = 0.72

(c) x_{A }= 0.4 ; x_{B} = 0.6

(d) x_{A} = 0.76 ; x_{B }= 0.24

Ans.

= 7 x 10

= (7 x 0.4 + 12 x 0.6) x 10

∴ x

(a) 136.8 g

(b) 17.1 g

(c) 68.4 g

(d) 34.2 g

Ans.

As we know,

So, no. of moles of sugar = 0.2 mole

Mass of sugar = No. of moles of sugar × Molar mass of sugar

= 0.2 × 342 = 68.4 g

(a) K

(b) K

(c) K

(d) K

Ans.

According to the question we can write

ΔT

ΔT

∴ K

∴ K

(a) 1 cup of water to 2 cups of pure milk

(b) 3 cups of water to 2 cups of pure milk

(c) 1 cup of water to 3 cups of pure milk

(d) 2 cups of water to 3 cups of pure milk

Ans.

Freezing point of diluted milk = -0.2°C

ΔT'

Freezing point of pure milk = -0.5°C

ΔT

∵ Moles of solute are same in both samples

2 cups of pure milk is mixed with 3 cups of water to make 5 cups of diluted milk.

(a) 1.6

(b) 1.8

(c) 2.0

(d) 2.2

Ans.

K

(a) 3A

(b) 2A

(c) A

(d) 4A

Ans.

(ΔT

K

m

(a) 0.2,22.20

(b) 0.2,11.11

(c) 0.167,11.11

(d) 0.167,22.20

Ans.

No. of moles of H

No. of moles of NaOH (n

**Q.15. Molecules of benzoic acid (C _{6}H_{5}COOH) dimerise in benzene, ‘w’ g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is:**(a)

(Given that K_{f} = 5 K kg mol^{-1}, Molar mass of benzoic acid = 122 g mol^{-1}) (2019)

(a) 2.4 g

(b) 1.0 g

(c) 1.5 g

(d) 1.8 g

Ans.

(a) 450 mm Hg, 0.4, 0.6

(b) 500 mm Hg, 0.5, 0.5

(c) 450 mm Hg, 0.5, 0.5

(d) 500 mm Hg, 0.4, 06

Ans.

= 0.5 × 600 + 0.5 × 400 = 300 + 200 = 500

Using the relation p

P

**Q.17. For the solution of the gases w, x,y and z in water at 298K, the Henry's law constants (K _{H}) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is: (2019)**

According to Henry’s law

y = c + mx

(x

x

y

y

(a)

(b)

(c)

(d)

Ans.

(a) 4 x 10

(b) 6 x 10

(c) 4 x 10

(d) 16 x 10

Ans.

we know,

∴ 2[XY] = 4(0.01)3

[XY] = 0.06

= 6 × 10

(Assume complete dissociation of the electrolyte) (2019)

(a) 0.18 K

(b) 0.24 K

(c) 0.12 K

(d) 0.36 K

Ans.

Dissociation of Potassium Sulphate (K

We know that, ΔT

where, K

= 3 × 4 × 0.03 = 0.36 K

(a) 1.08

(b) 1.35

(c) 1.48

(d) 1.51

Ans.

20% W/W KI solution (Given)

i.e. 100 g solution contains 20 g KI

∴ Mass of solvent = 100 - 20 = 80 g

(molar mass of urea = 60 g mol

(a) 0.027 mm Hg

(b) 0.028 mm Hg

(c) 0.017 mm Hg

(d) 0.031 mm Hg

Ans.

Relative lowering of vapour pressure, is given by,

Given,

Δp = pº - p = 0.017**Q.23. 1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, ΔT _{b}(A)/ΔT_{b}(B) is: (2019)**

Ebullioscopic constant (molal elevation const.) is given by,

(a) [CO(H

(b)[CO(H

(c) [CO(H

(d) [CO(H

Ans.

The complex giving least number of ions will have lowest depression in freezing point and therefore highest freezing point. Hence, option 1 is correct. (Van’t Hoff factor = 1)

(a) Freezing point of colloidal solution is lower than true solution at same concentration of a solute

(b) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed

(c) Colloidal particles can pass through ordinary filter paper

(d) When excess of electrolyte is added to colloidal solution, colloidal particle will be precipitated

Freezing point of colloidal solution is higher than true solution at same concentration of a solute.

(a) 37.5 g

(b) 75 g

(c) 150 g

(d) 50 g

Ans.

**Q.27. The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be****(K _{f} for benzene = 5.12 K kg mol^{–1}) (2017)**

⇒

α = 0.946

∴ % association = 94.6%

(K

(molar mass of S = 32 g mol

(a) 25 g

(b) 65 g

(c) 15 g

(d) 45 g

Ans.

(a) 0.162

(b) 0.675

(c) 0.325

(d) 0.486

Ans.

Molar mass of CHCl

Molar mass CH

Moles of CHCl

Moles of CH

Mole of CHCl

Mole fraction of CH

(Given - Vapour pressure of CHCl

Vapour pressure of CH

∴ P

= mole fraction of CHCl

= 0.5 × 0.263 + 0.5 × 0.546 = 0.4045

Mole fraction of CHCl

(a) 7.6

(b) 76.0

(c) 752.4

(d) 759.0

Ans.

According to Raoult's Law

Here P° = Vapour pressure of pure solvent,

P

W

M

Vapour pressure of pure water at 100° C (by assumption = 760 torr)

By substituting values in equation (i) we get,

On solving (ii) we get

P

(a) 0.02

(b) 0.015

(c) 0.0075

(d) 0.005

Ans.

According to Henry's law

S

(a) 0.67

(b) 0.33

(c) 0.80

(d) 0.50

Ans.

For MX

van 't Hoff factor (

⇒ α = 0.5

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