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Solutions · Question 13

JEE Main 2026 (Online) 24th January Morning ShiftMCQEasy📈 Topic in 26% papers+4−1
1

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

Answer

(1) 2.46 atm

Solution

Osmotic pressure depends on the total number of solute particles: = (nA + nB)VRT

nA = 0.360 = 0.005 mol, nB = 0.9180 = 0.005 mol

nA + nB = 0.01 mol; V = 100 mL = 0.1 L; T = 300 K

= 0.010.1 0.082 300

= 0.1 24.6 = 2.46 atm

Answer: (1)

JEE Main 2026 (Online) 24th January Morning Shift

Keep practising — Solutions

2

'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
3

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
4

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
5

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
6

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
7

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
8

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
9

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
10

The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is g L-1. (Nearest integer) Given : R = 0.08 L atm K-1 mol-1 Assume complete dissociation of NaCl (Given: Molar mass of Na and Cl are 23 and 35.5 g mol-1 respectively.)

JEE Main 2026 (Online) 21st January Evening Shift
11

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

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