JEE Main · 2025 · Shift-ImediumSOL-155

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass…

Solutions · Class 12 · JEE Main Previous Year Question

Question

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g mol−1^{-1}) and the decrease in freezing point is 0.40 K?

Options
  1. a

    5.125.12 K kg mol−1^{-1}

    ✓
  2. b

    4.434.43 K kg mol−1^{-1}

  3. c

    1.861.86 K kg mol−1^{-1}

  4. d

    3.723.72 K kg mol−1^{-1}

Correct Answera

5.125.12 K kg mol−1^{-1}

Detailed Solution

Strategy:\n> Rearrange the freezing point depression formula ΔTf=Kf⋅m\Delta T_f = K_f \cdot m to solve for KfK_f, given the mass of solute, its molar mass, and the mass of the solvent.\n\nStep 1: Calculate molality (mm)\n- textMolessolute=1textg/256textg/mol=0.003906textmol\\text{Moles solute} = 1\\text{ g} / 256\\text{ g/mol} = 0.003906\\text{ mol}.\n- textMasssolvent=50textg=0.05textkg\\text{Mass solvent} = 50\\text{ g} = 0.05\\text{ kg}.\nm=0.003906/0.05=0.078125textmol/kgm = 0.003906 / 0.05 = 0.078125\\text{ mol/kg}\n\nStep 2: Solve for } K_f \text{\nΔTf=0.40textK\Delta T_f = 0.40\\text{ K}.\n0.40=Kftimes0.078125  ⟹  Kf=0.40/0.078125=5.12textKkgmol−10.40 = K_f \\times 0.078125 \implies K_f = 0.40 / 0.078125 = 5.12\\text{ K kg mol}^{-1}\n(This corresponds to the value for benzene).\n\ntextAnswer:(1)\boxed{\\text{Answer: (1)}}

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