$1.00 \mathrm{~g}$ of non-eletrolyte solute (molar mass $250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved…

$1.00 \mathrm{~g}$ of non-eletrolyte solute (molar mass $250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $51.2 \mathrm{~g}$ of benzene. If the freezing point depression constant, $\mathrm{K}_f$ of benzene is 5.12 $\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$, the freezing point of benzene will be lowered by:
  1. $0.2 \mathrm{~K}$
  2. $0.4 \mathrm{~K}$
  3. $0.3 \mathrm{~K}$
  4. $0.5 \mathrm{~K}$

Solution

Molality of a non-electrolyte solute $\begin{aligned} & \frac{\frac{\text { weight of solute in gram }}{\text { molecular weight of solute }}}{\text { weight of solvent in kg }} \\ & =\frac{\frac{1}{250}}{0.0512}=\frac{1}{250 \times 0.0512} \\ & =0.0781 \mathrm{~m} \\ \Delta \mathrm{T}_f & =k_f \times \text { molality of solution } \\ & =5.12 \times 0.0781=0.4 \mathrm{~K} \end{aligned}$

Asked in: MHT CET Full Test 3

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