Integer Answer Type Questions for JEE: Solutions | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. At 300 K two liquids A and B forms an ideal solution. The vapour pressure of pure A and B are 400 mm and 600 mm respectively. If 1 mol of A and x moles of B are mixed, vapour pressure of solution becomes 550 mm. What is value of x

Ans. 3

∴ (1/(1+x)).400 + (x/(1+x)). 600
⇒ 400 + 600x = 550 + 550 x
⇒ 50 x = 150, x = 3

Q.2. The amount of urea to be dissolved in 500 ml of water (K = 1.86) to produce a depression of 1.86°C in freezing point is 10x gram. What is the value of x.

Ans. 3
ΔT_f = K_fm
⇒ 1.86 = 1.83 m ⇒ m = 1
For one molal solution amount of urea, CO(NH₂)₂, to be dissolved in 500 g H₂O is 30 gm
∴ 10x = 30 ⇒ x = 3

Q.3. Elevation in boiling point of a molar glucose solution (density = 1.2 g/ml) is  

Ans. 0.98 k_b

Molality of 1 < glucose solution =  0.98.

ΔT_b = k_b × m = 0.98 k_b.

Q.4. 0.04 M Na₂SO₄ solution is isotonic with 0.1 M glucose at the same temperature. What is the apparent degree of dissociation of Na₂SO₄? 

Ans. 0.75

At the same temperature, isotonic solutions will have same concentrations.

∴

0.04(1 + 2α) = 0.1
α = 0.75

Q.5. Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the  solution is (Atomic mass: H = 1, C = 12, O = 16)

Ans. 2.28 mol kg^–1

Molality =

= 2.28 mol kg^-1.

Q.6. Freezing point of 0.2 M KCN solution is –0.7°C. On adding 0.1 mole of Hg(CN)2 to one litre of the 0.2 M KCN solution, the freezing point of the solution becomes – 0.53°C  due to the reaction Hg(CN)₂ + mCN^– → Hg What is the value of m assuming molality = molarity?

Ans. 2

0.7 = 2 x K_f x 0.2 ⇒

Now, Molarity of K⁺ = 0.2 M
Molarity of CN^– = (0.2 – 0.1m) M
Molarity  of complex = 0.1 M
0.53 = K_f (0.2 + 0.2 – 0.1m + 0.1) = K_f (0.5 – 0.1m)
⇒ 0.53 = (7/4) (0.5 – 0.1m)
⇒ 2.1 = 3.5 – 0.7m
⇒ 0.7m = 1.4
⇒ m = 2.

Q.7. When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 747.3 Torr, the distillate contained 1.27 g of the liquid per gram of water. Calculate the molar mass of the liquid. The vapour pressure of water is 638.6 Torr at 95.2°C.

Ans. 134.3 h mol^-1

For a mixture of two immiscible liquids, the total pressure is given by

 ...(i)
and  ...(ii)
Dividing equation (i) by (ii) gives,

or
where n_l and n_w represents the number of moles of liquid and water in vapour phase, respectively.
 

where M_l and  represents molar mass of liquid and water respectively and w_l and  denotes weight of liquid and water in vapour phase, respectively.

 

Q.8. 1000 gm of sucrose solution in water is cooled to –0.5°C. How much of ice would be separated out at this temperature, if the solution started to freeze at –0.38°C. Express your answer is gram. (K_f H₂O = 1.86 K mol^–1 kg ).

Ans. 0223 gm

w_A + w_B = 1000 …(i)

or
w_B/w_A = 0.07 …(ii)
By (i) and (ii), w_B = 65.42 gm and w_A =  934.58 gm
At –0.5°C, some water separated out as ice and solute exist as 65.42 gm.

w_A =  711.58 gm
∴ separated ice = 934.58 gm – 711.58 gm = 0223 gm.

Q.9. What weight of solute (molecular weight 60) is required to dissolve in 180 g of water to reduce the vapour pressure to 4/5th pure water?

Ans. 150 g.

 

P_s = 4P°/5, m = 60,  w = ?,  W = 180 g,  M = 18

Q.10. The freezing point of 0.08m NaHSO₄ is –0.345°C. Calculate the percentage of that transfer a proton to water to form SO^-2₄ ion.
K_f of H₂O = 1.86 mol^–1 kg.

Ans. 66%

NaHSO₄ +  H₂O(aq) → Na⁺ +  H₃O⁺ +  SO₄^–2

ΔT_f = i × K_f × m
0.345 = (1 + 2α) × 1.86 × 0.08
So, α = 0.6592
% dissociation = 65.92%