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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2019 · 10 January · Morning Shift · Chemistry · Solutions

Liquids A and B form an ideal solution in the entire composition...

Liquids AA and BB form an ideal solution in the entire composition range. At 350K,350K, the vapour pressure of pure A and pure B are 7×103Pa7 \times {10}^{3}Pa and 12×103Pa12 \times {10}^{3}Pa , respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is:

Options

  1. AxA=0.4;xB=0.6{x}_{A} = 0.4; {x}_{B} = 0.6
  2. BxA=0.76;xB=0.24{x}_{A} = 0.76; {x}_{B} = 0.24
  3. CxA=0.28;xB=0.72{x}_{A} = 0.28; {x}_{B} = 0.72
  4. DxA=0.37;xB=0.63{x}_{A} = 0.37; {x}_{B} = 0.63

Correct answer

C. xA=0.28;xB=0.72{x}_{A} = 0.28; {x}_{B} = 0.72

Step-by-step solution

Since, PA0=7×103Pa;PB0=12×103Pa{P}_{A}^{0} = 7 \times {10}^{3}Pa; {P}_{B}^{0} = 12 \times {10}^{3}Pa and XA=0.4{X}_{A} = 0.4, XB=1−0.4=0.6{X}_{B} = 1 - 0.4 = 0.6

According to Raoult's Law: PT=PA0XA+PB0XB{P}_{T} = {P}_{A}^{0}{X}_{A} + {P}_{B}^{0}{X}_{B}

PT=(7×103)(0.4)+(12×103)(0.6){P}_{T} = \left(7 \times {10}^{3}\right)\left(0.4\right) + \left(12 \times {10}^{3}\right)\left(0.6\right)

PT=10×103Pa{P}_{T} = 10 \times {10}^{3}Pa

Now, PA0XA=PTxA{P}_{A}^{0}{X}_{A} = {P}_{T}{x}_{A}

xA=PA0XAPT=7×103×0.4104⇒0.28{x}_{A} = \dfrac{{P}_{A}^{0}{X}_{A}}{{P}_{T}} = \dfrac{7 \times {10}^{3} \times 0.4}{{10}^{4}} ⇒ 0.28

xB=1−xA⇒1−0.28⇒0.72{x}_{B} = 1 - {x}_{A} ⇒ 1 - 0.28 ⇒ 0.72

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