JEE Main · 2019 · Shift-ImediumSOL-066

Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressure of pure A and…

Solutions · Class 12 · JEE Main Previous Year Question

Question

Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressure of pure A and pure B are 7×1037 \times 10^3 Pa and 12×10312 \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. a

    xA=0.4; xB=0.6x_A = 0.4;\ x_B = 0.6

  2. b

    xA=0.76; xB=0.24x_A = 0.76;\ x_B = 0.24

  3. c

    xA=0.28; xB=0.72x_A = 0.28;\ x_B = 0.72

    ✓
  4. d

    xA=0.37; xB=0.63x_A = 0.37;\ x_B = 0.63

Correct Answerc

xA=0.28; xB=0.72x_A = 0.28;\ x_B = 0.72

Detailed Solution

Strategy:\n> Use Raoult's law to find partial pressures from liquid composition, then use Dalton's law to find the mole fraction of each component in the resulting vapour.\n\nStep 1: Define liquid phase concentrations\n- xA=0.40x_A = 0.40\n- xB=0.60x_B = 0.60\n\nStep 2: Calculate partial pressures\n- PA=xAtimesPA∘=0.40times7000textPa=2800textPaP_A = x_A \\times P_A^\circ = 0.40 \\times 7000\\text{ Pa} = 2800\\text{ Pa}\n- PB=xBtimesPB∘=0.60times12000textPa=7200textPaP_B = x_B \\times P_B^\circ = 0.60 \\times 12000\\text{ Pa} = 7200\\text{ Pa}\n\nStep 3: Calculate total pressure\nPtexttotal=2800+7200=10000textPaP_{\\text{total}} = 2800 + 7200 = 10000\\text{ Pa}\n\nStep 4: Find vapour phase mole fractions (yy)\n- yA=fracPAPtexttotal=frac280010000=0.28y_A = \\frac{P_A}{P_{\\text{total}}} = \\frac{2800}{10000} = 0.28\n- yB=fracPBPtexttotal=frac720010000=0.72y_B = \\frac{P_B}{P_{\\text{total}}} = \\frac{7200}{10000} = 0.72\n\ntextAnswer:(3)\boxed{\\text{Answer: (3)}}

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