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Solutions question

2022 · 28 Jul · Shift 2 · Q16
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Solutions question

2022 · 28 Jul · Shift 2 · Q16

JEE MainChemistrySolutionsNumerical+4 / −1
A gaseous mixture of two substances A and B, under a total pressure of 0.80.80.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.50.50.5 in the vapour phase and 0.20.20.2 in the liquid phase. The vapour pressure of pure liquid A\mathrm{A}A is ‾\underline{\hspace{2cm}}​ atm. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 2

  1. Given data
  • Total pressure of vapour mixture: P=0.8 atmP = 0.8\,\text{atm}P=0.8atm
  • Mole fraction of AAA in vapour phase: yA=0.5y_A = 0.5yA​=0.5
  • Mole fraction of AAA in liquid phase: xA=0.2x_A = 0.2xA​=0.2
  • The liquid solution is ideal

We need to find vapour pressure of pure liquid AAA, i.e. PA0P_A^0PA0​.


  1. Use Dalton’s law to find partial pressure of AAA in vapour

For the vapour phase, pA=yAPp_A = y_A PpA​=yA​P

So, pA=0.5×0.8=0.4 atmp_A = 0.5 \times 0.8 = 0.4\,\text{atm}pA​=0.5×0.8=0.4atm


  1. Use Raoult’s law for an ideal solution

For component AAA, pA=xAPA0p_A = x_A P_A^0pA​=xA​PA0​

Substitute the values: 0.4=0.2×PA00.4 = 0.2 \times P_A^00.4=0.2×PA0​

Therefore, PA0=0.40.2=2 atmP_A^0 = \frac{0.4}{0.2} = 2\,\text{atm}PA0​=0.20.4​=2atm


  1. Nearest integer

PA0=2 atmP_A^0 = 2\,\text{atm}PA0​=2atm

So the required integer answer is: 2\boxed{2}2​


  1. Comparison with stored answer

Stored correct answer = 222

Our derived answer also = 222

Hence, the answer agrees with the stored correct answer.

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