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A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of…

Solutions · Class 12 · JEE Main Previous Year Question

Question

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

Options
  1. a

    1400 mm Hg, A

  2. b

    1400 mm Hg, B

  3. c

    600 mm Hg, B

  4. d

    600 mm Hg, A

    ✓
Correct Answerd

600 mm Hg, A

Detailed Solution

Strategy:\n> Apply Raoult's Law PT=xAPA∘+xBPB∘P_T = x_A P_A^\circ + x_B P_B^\circ to find the unknown vapor pressure of component B.\n\nStep 1: Calculate mole fractions\nnA=1,nB=3n_A = 1, n_B = 3.\n- xA=1/4=0.25x_A = 1/4 = 0.25.\n- xB=3/4=0.75x_B = 3/4 = 0.75.\n\nStep 2: Solve Raoult's Equation\nPT=500,PA∘=200P_T = 500, P_A^\circ = 200.\n500=(0.25times200)+(0.75timesPB∘)500 = (0.25 \\times 200) + (0.75 \\times P_B^\circ)\n500=50+0.75PB∘500 = 50 + 0.75 P_B^\circ\n450=0.75PB∘  ⟹  PB∘=600textmmHg450 = 0.75 P_B^\circ \implies P_B^\circ = 600\\text{ mmHg}\n\nStep 3: Identify least volatile\nPA∘=200P_A^\circ = 200 and PB∘=600P_B^\circ = 600.\nA has lower vapor pressure impliesimplies A is least volatile.\n\ntextAnswer:(4)\boxed{\\text{Answer: (4)}}

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