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Solutions · Question 11

JEE Main 2026 (Online) 28th January Morning ShiftMCQMedium📈 Topic in 16% papers+4−1
1

At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:

  1. 400, 300
  2. 600, 400
  3. 500, 200
  4. 300, 200

Answer

(3) 500, 200

Solution

For an ideal solution: Ptotal = pA0xA + pB0xB

Solution 1: xA = 25, xB = 35

25pA0 + 35pB0 = 320 2pA0 + 3pB0 = 1600 … (i)

Solution 2 (3 mol A, 4 mol B): xA = 37, xB = 47

37pA0 + 47pB0 = 328.6 3pA0 + 4pB0 = 2300.2 … (ii)

3 × (i) − 2 × (ii): 9pB0 - 8pB0 = 4800 - 4600.4

pB0 = 199.6 200 mm Hg

From (i): 2pA0 = 1600 - 598.8 = 1001.2 pA0 = 500.6 500 mm Hg

Answer: (3)

JEE Main 2026 (Online) 28th January Morning Shift

Keep practising — Solutions

2

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]

  1. III < I < II < IV
  2. II < III < IV < I
  3. II < III < I < IV
  4. I < II < III < IV
JEE Main 2026 (Online) 28th January Evening Shift
3

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass 60 g mol-1 and 0.9 g of a non-volatile non-electrolyte solute 'B' of molar mass 180 g mol-1 in 100 mL H2O at 27 . Osmotic pressure of the solution will be [Given: R = 0.082 L atm K-1 mol-1]

  1. 2.46 atm
  2. 0.82 atm
  3. 1.47 atm
  4. 1.23 atm
JEE Main 2026 (Online) 24th January Morning Shift
4

'W' g of a non-volatile electrolyte solid solute of molar mass 'M' g mol-1 when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and Kb for water is 0.52 K kg mol-1, then the mole fraction of the electrolyte solute (x2) in the solution can be expressed as (Given : density of water = 1 g/mL and boiling point of water = 373 K)

  1. 162.6 WM
  2. 1.38 MW
  3. 2.616 MW
  4. 1.38 WM
JEE Main 2026 (Online) 24th January Morning Shift
5

At 298 K, the mole percentage of N2(g) in air is 80\%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N2(g) in water at 298 K? (KH for N2 is 6.5 107 mmHg)

  1. 9.35 10-5
  2. 9.35 105
  3. 1.23 10-7
  4. 1.17 10-4
JEE Main 2026 (Online) 24th January Evening Shift
6

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m-2 respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

  1. 0.5217
  2. 0.480
  3. 0.340
  4. 0.663
JEE Main 2026 (Online) 24th January Evening Shift
7

Which one of the following graphs accurately represents the plot of partial pressure of CS2 vs its mole fraction in a mixture of acetone and CS2 at constant temperature?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 23rd January Morning Shift
8

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is . (Nearest integer)

JEE Main 2026 (Online) 23rd January Evening Shift
9

Consider a solution of CO2(g) dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO2 in vapour phase above water) [y-axis] with log (mole fraction of CO2 in water) [x-axis] at 25 ?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 22nd January Morning Shift
10

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below

  1. Both Statement I and Statement II are true
  2. Statement I is true but Statement II is false
  3. Statement I is false but Statement II is true
  4. Both Statement I and Statement II are false
JEE Main 2026 (Online) 22nd January Morning Shift
11

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2. When 1 g of PQ is dissolved in 50 g of solvent 'A', Tb was 1.176 K while when 1 g of PQ2 is dissolved in 50 g of solvent 'A', Tb was 0.689 K. (Kb of 'A' = 5 K kg mol-1). The molar masses of elements P and Q (in g mol-1) respectively, are :

  1. 25, 60
  2. 60, 25
  3. 65, 145
  4. 70, 110
JEE Main 2026 (Online) 21st January Morning Shift

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