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Solutions · Question 33

JEE Main 2025 (Online) 2nd April Morning ShiftMCQEasy📈 Topic in 16% papers+4−1
1

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

  1. 1400 mm Hg, A
  2. 1400 mm Hg, B
  3. 600 mm Hg, A
  4. 600 mm Hg, B

Answer

(3) 600 mm Hg, A

Solution

Mole fractions: xA = 11+3 = 0.25, xB = 34 = 0.75

By Raoult's law, ptotal = pA0 xA + pB0 xB

500 = 200 0.25 + pB0 0.75

0.75 pB0 = 450

pB0 = 600 mm Hg

pA0 = 200 mm Hg < pB0 = 600 mm Hg, A is the least volatile component.

Answer: (3)

JEE Main 2025 (Online) 2nd April Morning Shift

Keep practising — Solutions

2

Which of the following graph correctly represents the plots of KH at 1 bar for gases in water versus temperature?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 2nd April Morning Shift
3

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 10-1 (Nearest integer) (Given : Molal boiling point elevation constant is 0.5 K kg mol-1)

JEE Main 2025 (Online) 2nd April Evening Shift
4

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 100.0156°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?

  1. AX2 and AY2 (both) are completely unionised.
  2. AX2 and AY2 (both) are fully ionised.
  3. AX2 is completely unionised while AY2 is fully ionised.
  4. AX2 is fully ionised while AY2 is completely unionised.
JEE Main 2025 (Online) 29th January Morning Shift
5

If A2B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is 10-1.

JEE Main 2025 (Online) 29th January Morning Shift
6

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol-1) and the decrease in freezing point is 0.40 K?

  1. 4.43 K kg mol-1
  2. 3.72 K kg mol-1
  3. 5.12 K kg mol-1
  4. 1.86 K kg mol-1
JEE Main 2025 (Online) 28th January Morning Shift
7

Assume a living cell with 0.9% (w/w) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will :

  1. shrink since solution is 0.45% (w/w) as a result of association of glucose molecules (due to hydrogen bonding)
  2. shrink since solution is 0.5% (w/w)
  3. show no change in volume since solution is 0.9% (w/w)
  4. swell up since solution is 1% (w/w)
JEE Main 2025 (Online) 28th January Evening Shift

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