JEE  >  Integer Answer Type Questions: Solutions

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

Document Description: Integer Answer Type Questions: Solutions for JEE 2022 is part of All Types of Questions for JEE preparation. The notes and questions for Integer Answer Type Questions: Solutions have been prepared according to the JEE exam syllabus. Information about Integer Answer Type Questions: Solutions covers topics like and Integer Answer Type Questions: Solutions Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Integer Answer Type Questions: Solutions.

Introduction of Integer Answer Type Questions: Solutions in English is available as part of our All Types of Questions for JEE for JEE & Integer Answer Type Questions: Solutions in Hindi for All Types of Questions for JEE course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE
 1 Crore+ students have signed up on EduRev. Have you?

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 =  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.

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 Kf 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 = 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

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

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

where Mand  represents molar mass of liquid and water respectively and wl 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. (Kf H2O = 1.86 K mol–1 kg ).

Ans. 0223 gm

wA + wB = 1000 …(i)

or
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.

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.

Ps = 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.
Kf of H2O = 1.86 mol–1 kg.

Ans. 66%

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

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

The document Integer Answer Type Questions: Solutions Notes | Study All Types of Questions for JEE - JEE is a part of the JEE Course All Types of Questions for JEE.
All you need of JEE at this link: JEE

All Types of Questions for JEE

376 docs
 Use Code STAYHOME200 and get INR 200 additional OFF

All Types of Questions for JEE

376 docs

Top Courses for JEE

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;