JEE Main · 2019 · Shift-IIeasySOL-108

The elevation in boiling point for 1 molal solution of glucose is 2 K. The depression in freezing point for 2 molal…

Solutions · Class 12 · JEE Main Previous Year Question

Question

The elevation in boiling point for 1 molal solution of glucose is 2 K. The depression in freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between KbK_b and KfK_f is:

Options
  1. a

    Kb=2KfK_b = 2K_f

  2. b

    Kb=1.5KfK_b = 1.5K_f

  3. c

    Kb=KfK_b = K_f

  4. d

    Kb=0.5KfK_b = 0.5K_f

Correct Answera

Kb=2KfK_b = 2K_f

Detailed Solution

Strategy:\n> Use the definitions of ΔTb\Delta T_b and ΔTf\Delta T_f to solve for the constants KbK_b and KfK_f, then determine the mathematical relationship between them.\n\nStep 1: Determine Kb\nFor 1 molal glucose: ΔTb=2textK\Delta T_b = 2\\text{ K}.\n2=Kbtimes1    Kb=2textKkgmol12 = K_b \\times 1 \implies K_b = 2\\text{ K kg mol}^{-1}\n\nStep 2: Determine Kf\nFor 2 molal glucose: ΔTf=2textK\Delta T_f = 2\\text{ K}.\n2=Kftimes2    Kf=1textKkgmol12 = K_f \\times 2 \implies K_f = 1\\text{ K kg mol}^{-1}\n\nStep 3: Compare results\nKb=2K_b = 2 and Kf=1K_f = 1. Thus Kb=2KfK_b = 2 K_f.\n\ntextAnswer:(1)\boxed{\\text{Answer: (1)}}

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