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Solutions question

2022 · 28 Jun · Shift 1 · Q18
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Solutions question

2022 · 28 Jun · Shift 1 · Q18

JEE MainChemistrySolutionsNumerical+4 / −1
The vapour pressures of two volatile liquids A and B at 25 ∘^\circ∘ C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is x17{x \over {17}}17x​. The value of x is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 14

  1. Given data
  • Vapour pressure of pure liquid A: PA0=50P_A^0 = 50PA0​=50 Torr
  • Vapour pressure of pure liquid B: PB0=100P_B^0 = 100PB0​=100 Torr
  • Mole fraction of A in liquid phase: xA=0.3x_A = 0.3xA​=0.3

So, mole fraction of B in liquid phase is xB=1−xA=1−0.3=0.7x_B = 1 - x_A = 1 - 0.3 = 0.7xB​=1−xA​=1−0.3=0.7

  1. Use Raoult’s law to find partial pressures

For an ideal volatile liquid mixture, PA=xAPA0=0.3×50=15 TorrP_A = x_A P_A^0 = 0.3 \times 50 = 15 \text{ Torr}PA​=xA​PA0​=0.3×50=15 Torr PB=xBPB0=0.7×100=70 TorrP_B = x_B P_B^0 = 0.7 \times 100 = 70 \text{ Torr}PB​=xB​PB0​=0.7×100=70 Torr

  1. Find total vapour pressure

Ptotal=PA+PB=15+70=85 TorrP_{\text{total}} = P_A + P_B = 15 + 70 = 85 \text{ Torr}Ptotal​=PA​+PB​=15+70=85 Torr

  1. Find mole fraction of B in vapour phase

If yBy_ByB​ is the mole fraction of B in vapour phase, then yB=PBPtotal=7085=1417y_B = \frac{P_B}{P_{\text{total}}} = \frac{70}{85} = \frac{14}{17}yB​=Ptotal​PB​​=8570​=1714​

  1. Compare with given form

Given that mole fraction of liquid B in vapour phase is x17\dfrac{x}{17}17x​, so x17=1417\frac{x}{17} = \frac{14}{17}17x​=1714​

Hence, x=14x = 14x=14

  1. Comparison with stored answer

Stored correct answer = 14

Our derived answer also is 14, so it agrees.

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