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Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE

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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
ΔTf = Kfm
⇒ 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 kb

Molality of 1 < glucose solution = Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE 0.98.

ΔTb = kb × m = 0.98 kb.


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.

Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
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 = Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE

= 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 → HgInteger Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE What is the value of m assuming molality = molarity?

Ans. 2

0.7 = 2 x Kf x 0.2 ⇒ Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
Now, Molarity of K+ = 0.2 M
Molarity of CN = (0.2 – 0.1m) M
Molarity  of complex = 0.1 M
0.53 = Kf (0.2 + 0.2 – 0.1m + 0.1) = Kf (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

Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE ...(i)
and Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE ...(ii)
Dividing equation (i) by (ii) gives,
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
or Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
where nl and nw represents the number of moles of liquid and water in vapour phase, respectively.
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE 
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
where Mand Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE represents molar mass of liquid and water respectively and wl and Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE denotes weight of liquid and water in vapour phase, respectively.
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE


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. (Kf H2O = 1.86 K mol–1 kg ).

Ans. 0223 gm

wA + wB = 1000 …(i)
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
or Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
wB/wA = 0.07 …(ii)
By (i) and (ii), w= 65.42 gm and wA =  934.58 gm
At –0.5°C, some water separated out as ice and solute exist as 65.42 gm.
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
wA =  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.

 Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE

Ps = 4P°/5, m = 60,  w = ?,  W = 180 g,  M = 18
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE


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.
Kf of H2O = 1.86 mol–1 kg.

Ans. 66%

NaHSO4 +  H2O(aq) → Na+ +  H3O+ +  SO4–2
Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
ΔTf = i × Kf × m
0.345 = (1 + 2α) × 1.86 × 0.08
So, α = 0.6592
% dissociation = 65.92%  Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE

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