Q11 / 97 · NTA Abhyas (JEE Main) 2020 · Chemistry · Solutions

The vapour pressures of pure liquids A and B are 400 and 600 mmHg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fractional of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are

  • A
    500 mmHg, 0.5, 0.5
  • B
    450 mmHg, 0.4, 0.6
  • C
    450 mmHg, 0.5, 0.5
  • Correct answer
    D
    500 mmHg, 0.4, 0.6

Solution

Let XA and XB be the mole fraction of liquid A and B in the mixture.

Given, PA°=400 mm Hg, PB°=600 mm Hg

P Total = X A . P A ° + X B . P B °
=0.5×400+0.5×600
=500 mm Hg

Now, mole fraction of A in vapour,

YA=PAPtotal=0.5×400500=0.4 ∵[PA=XAPA°]

and mole fraction of B in vapour, YB=1-0.4=0.6

Hence, the correct answer is option: 'D'.

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