Topic wise assignment class 12 chemistry

 Topic : Raoult’s Law

1. Mixture (s) showing positive deviation from Raoult’s law at 35oC is (are)    JEE Advanced 2016 Paper 2 Offline

(a) carbon tetrachloride + methanol

(b) carbon disulphide + acetone

(c) benzene + toluene

(d) phenol + aniline

Ans : a and b both        

2. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm–3 . The ratio of the molecular weights of the solute and solvent, , is                          JEE Advanced 2016 Paper 1 Offline

Answer is 9                                            

3. The Henry's law constant for the solubility of N gas in water at 298 K is 1.0  10 atm. The mole fraction of N in air is 0.8. The number of moles of N from air dissolved in 10 moles of water at 298 K and 5 atm pressure is                                            IIT-JEE 2009 Paper 1 Offline

(a) 4.0 x 10-4

(b) 4.0 x 10-5

(c) 5.0 x 10-4

(d) 4.0 x 10-6

Ans: 4.0 x 10-4

4. The vapour pressure of ethanol and methanol are 44.5 and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.                                                                                        IIT-JEE 1986

Answer is 66.17 mm, 0.65

Topic : Elevation in Boiling Point

1. Consider the following aqueous solutions.                           JEE Main 2026 (Online) 28th January Evening Shift

I. 2.2 g Glucose in 125 mL of solution.

II. 1.9 g Calcium chloride in 250 mL of solution.

III. 9.0 g Urea in 500 mL of solution.

IV. 20.5 g Aluminium sulphate in 750 mL of solution.

The correct increasing order of boiling point of these solutions will be :

[Given : Molar mass in g mol−1 : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]

(a) III < I < II < IV

(b) II < III < IV < I

(c) II < III < I < IV

(d) I < II < III < IV

Ans : I < II < III < IV\

2. A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol−1) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is ______ × 10−2. (nearest integer)

[Given: Kb of the solvent = 5.0 K kg mol−1]

Assume the solution to be dilute and no association or dissociation of X takes place in solution.                                                                JEE Main 2026 (Online) 21st January Evening Shift

Ans: 3

3. Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and . When 1 g of  is dissolved in 50 g of solvent '  was 1.176 K while when 1 g of  is dissolved in 50 g of solvent '  was 0.689 K . ( Kb of 'A` = 5 K kg/mol). The molar masses of elements P and Q (in g/mol ) respectively, are :                                                JEE Main 2026 (Online) 21st January Morning Shift

(a)  25,60     (b) 60,25     (c) 65,145     (d) 70,110

Ans: (a) 25,60 

4. 2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water is 0.52 K kg/mol) 

(a) 377.3 K    (b) 379.2 K     (c) 375.3 K    (d) 277.3 K        JEE Main 2025 (Online) 3rd April Morning Shift

Ans: (a) 377.3 K 

5. When 1 g each of compounds AB and AB2 are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu ) is_________  (Nearest integer)                                

JEE Main 2025 (Online) 2nd April Evening Shift

(Given : Molal boiling point elevation constant is  )

Ans: 2.5

6. 1.24 g of AX2 (molar mass 124 g mol−1) is dissolved in 1 kg of water to form a solution with boiling point of 100.015°C, while 25.4 g of AY2 (molar mass 250 g mol−1) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.

Kb(H2O) = 0.52 kg mol−1

Which of the following is correct?                                JEE Main 2025 (Online) 29th January Morning Shift

(a) AX2 and AY2 (both) are completely unionised.

(b) AX2 and AY2 (both) are fully ionised.

(c) AX2 is completely unionised while AY2 is fully ionised.

(d) AX2 is fully ionised while AY2 is completely unionised.

Ans: (d) AX2 is fully ionised while AY2 is completely unionised.

7. Arrange the following solutions in order of their increasing boiling points.

(i) 10-4 M NaCl

(ii) 10-4 M Urea

(iii) 10-3 M NaCl

(iv) 10-2 M NaCl                                        JEE Main 2025 (Online) 22nd January Morning Shift

(a) i<ii<iii< iv

(b) ii < i < iii < iv

(c) iv < iii < i < ii

(d) ii < i = iii <iv

8. A solution containing 10g of an electrolyte AB2 in 100 g of water boils at100.52 C. The degree of ionization of the electrolyte 

 is _________ . (nearest integer)

[Given : Molar mass of  (molal boiling point elevation const. of water) , boiling point of water  ionises as

AB2 --> A2+ + 2B-                                                              JEE Main 2024 (Online) 8th April Morning Shift

Answer = 5

9.  2.5g of a non-volatile, non-electrolyte is dissolved in 100g of water at 25 C. The solution showed a boiling point elevation by 2 C. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ mm of Hg (nearest integer)                            JEE Main 2024 (Online) 4th April Morning Shift

[Given : Molal boiling point elevation constant of water ,  pressure  of , molar mass of water 

Ans : the vapor pressure of the aqueous solution is 707 mm Hg.

10. A solution of two miscible liquids showing negative deviation from Raoult's law will have

(a) increased vapour pressure, increased boiling point

(b) increased vapour pressure, decreased boiling point

(c) decreased vapour pressure, decreased boiling point

(d) decreased vapour pressure, increased boiling point 

JEE Main 2024 (Online) 27th January Morning Shift

Ans:  (d) decreased vapour pressure, increased boiling point

Topic : Depression in Freezing Point

1. On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decreases from 650 mmHg to 640 mmHg. The depression of freezing point of benzene (in K) upon addition of the solute is .............

(Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol-1 and 5.12 K kg mol-1, respectively).                                JEE Advanced 2019 Paper 1 Offline

Answer = 1.02

2. Pure water freezes at   and  bar. The addition of   of ethanol to   of water changes the freezing point of the solution. Use the freezing point depression constant of water as  kg  The figures shown below represent plots of vapor pressure  versus temperature  [molecular weight of ethanol is    ] Among the following, the option representing change in the freezing point is                    JEE Advanced 2017 Paper 2 Offline

(a) 

(b) 

(c) 

(d) 

3.If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex(which behaves as a strong electrolyte) is – 0.0558oC, the number of chloride(s) in the coordination sphere of the complex is [Kf of water = 1.86 K kg mol–1 ]        

Answer = 1                                                                                JEE Advanced 2015 Paper 1 Offline

4. Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.

Given:

Freezing point depression constant of water  K kg mol

Freezing point depression constant of ethanol  K kg mol

Boiling point elevation constant of water  K kg mol

Boiling point elevation constant of ethanol  K kg mol

Standard freezing point of water = 273 K

Standard freezing point of ethanol = 155.7 K

Standard boiling point of water = 373 K

Standard boiling point of ethanol = 351.5 K

Vapour pressure of pure water = 32.8 mm Hg

Vapour pressure of pure water = 40 mm Hg

Molecular weight of water = 18 g mol

Molecular weight of ethanol = 46 g mol

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.                           IIT-JEE 2008 Paper 1 Offline

The freezing point of the solution M is :

(a) 268.7 K            (b) 268.5 K        (c) 234.2 K            (d) 150.9 K

Answer = (d) 150.9 K

5. 75.2 g of C6H5OH (phenol) is dissolved in a solvent of Kf = 14. If the depression in freezing point is 7 K then find the % of phenol that dimerises.                            IIT-JEE 2006

Answer = 75

6. A solution of a nonvolatile solute in water freezes at -0.30oC. The vapour pressure of pure water at 298 K s 23.51 mm Hg and Kf for water is 1.86 K kg mol-1. Calculate the vapour pressure of this solution at 298 K.                                                          IIT-JEE 1998

Answer is 23.44 mm Hg

7. The freezing point of equimolal aqueous solutions will be highest for        IIT-JEE 1990

(a) C6H5NH3Cl (aniline hydrochloride)

(b)  Ca(NO3)2

    (c) La(NO3)3

      (d) C6H12O6 (glucose)

Answer : (d) C6H12O6 (glucose)