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JEE Main 2025 (Online) 23rd January Evening ShiftMCQEasy📈 Topic in 3% papers+4−1
Ques. 1JEE Main 2025, 23 January (Evening Shift)

When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?

  1. 0.2
  2. 0.4
  3. 0.8
  4. 0.6

Answer

(4) 0.6

Solution

p = p0 xsolute

10 = p0 0.2 p0 = 50 mm Hg

20 = 50 xsolute xsolute=0.4

∴ xsolvent=1-0.4=0.6

Answer: (4)

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Ques. 2JEE Main 2025, 23 January (Evening Shift)

Consider a binary solution of two volatile liquid components 1 and 2. x1 and y1 are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1x1 vs 1y1 are given respectively as :

  1. P10P20, P10-P20P20
  2. P20P10, P20-P10P10
  3. P20P10, P10-P20P20
  4. P10P20, P20-P10P20
Ques. 3JEE Main 2025, 22 January (Morning Shift)

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

  1. (i) < (ii) < (iii) < (iv)
  2. (ii) < (i) < (iii) < (iv)
  3. (iv) < (iii) < (i) < (ii)
  4. (ii) < (i) (iii) < (iv)
Ques. 4JEE Main 2025, 22 January (Evening Shift)

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is :

  1. 2.79 m
  2. 2 m
  3. 1.79 m
  4. 3 m

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