What is the depression of freezing point, when mole fraction of non-electrolyte solute in aqueous solution…

What is the depression of freezing point, when mole fraction of non-electrolyte solute in aqueous solution is $0.01 ?\left(\mathrm{~K}_{\mathrm{f}}\right.$ of $\left.\mathrm{H}_2 \mathrm{O}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
  1. $1.246 \mathrm{~K}$
  2. $1.380 \mathrm{~K}$
  3. $1.528 \mathrm{~K}$
  4. $1.043 \mathrm{~K}$

Solution

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{m} \times \mathrm{K}_{\mathrm{f}}=\frac{\mathrm{n}_2}{\mathrm{~m}_1} \times \mathrm{K}_{\mathrm{f}}$ We have, $\mathrm{x}_2=\frac{\mathrm{n}_2}{\mathrm{n}_1+\mathrm{n}_2} \approx \frac{\mathrm{n}_2}{\mathrm{n}_1}=0.01$ Taking $1 \mathrm{~L}$ of water, mass of water $=1 \mathrm{~kg}=1000 \mathrm{~g}$ $ \begin{aligned} & \Rightarrow \mathrm{n}\left(\mathrm{H}_2 \mathrm{O}\right)=\mathrm{n}_1=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=55.55 \mathrm{~mol} \\ & \Rightarrow \mathrm{n}_2=55.55 \times 0.01=0.556 \mathrm{~mol} \\ & \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}=\frac{\mathrm{n}_2}{\mathrm{~m}_1} \times \mathrm{K}_{\mathrm{f}}=\frac{0.556 \mathrm{~mol}}{1 \mathrm{~kg}} \times 1.86 \\ & =1.034 \mathrm{~K} \approx 1.043 \mathrm{~K} \end{aligned} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

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