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.

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.

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 . (  of '

 ). The molar masses of elements P and Q (in  ) respectively, are :

(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  )

(a) 377.3 K    (b) 379.2 K     (c) 375.3 K    (d) 277.3 K

Ans: (a) 377.3 K 

5. When 1 g each of compounds AB and  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  ) is____________  (Nearest integer)

(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?

(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.

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)