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Solutions question

2025 · 28 Jan · Shift 1 · Q2
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Solutions question

2025 · 28 Jan · Shift 1 · Q2

JEE MainChemistrySolutionsMCQ+4 / −1
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−1256 \mathrm{~g} \mathrm{~mol}^{-1}256 g mol−1 ) and the decrease in freezing point is 0.40 K ?
  1. A
    4.43 K kg mol−14.43 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}4.43 K kg mol−1
  2. B
    3.72 K kg mol−13.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}3.72 K kg mol−1
  3. C
    5.12 K kg mol−15.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}5.12 K kg mol−1
  4. D
    1.86 K kg mol−11.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}1.86 K kg mol−1
View written solutionFree

Correct answer: C

  1. Use the freezing point depression formula

    ΔTf=Kf⋅m\Delta T_f = K_f \cdot mΔTf​=Kf​⋅m

    where:

    • ΔTf=0.40 K\Delta T_f = 0.40\,\text{K}ΔTf​=0.40K
    • Kf=K_f =Kf​= freezing point depression constant
    • m=m =m= molality

  2. Calculate moles of solute

    Given:

    • mass of solute =1 g= 1\,\text{g}=1g
    • molar mass =256 g mol−1= 256\,\text{g mol}^{-1}=256g mol−1

    n=1256=0.00390625 moln = \frac{1}{256} = 0.00390625\,\text{mol}n=2561​=0.00390625mol

  3. Calculate mass of solvent in kg

    Given solvent mass =50 g=0.050 kg= 50\,\text{g} = 0.050\,\text{kg}=50g=0.050kg

  4. Calculate molality

    m=0.003906250.050=0.078125 mol kg−1m = \frac{0.00390625}{0.050} = 0.078125\,\text{mol kg}^{-1}m=0.0500.00390625​=0.078125mol kg−1

  5. Calculate KfK_fKf​

    Kf=ΔTfm=0.400.078125=5.12 K kg mol−1K_f = \frac{\Delta T_f}{m} = \frac{0.40}{0.078125} = 5.12\,\text{K kg mol}^{-1}Kf​=mΔTf​​=0.0781250.40​=5.12K kg mol−1

  6. Match with the options

    Kf=5.12 K kg mol−1K_f = 5.12\,\text{K kg mol}^{-1}Kf​=5.12K kg mol−1

    So the correct option is C.

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