JEE MAIN - Chemistry (2023 - 1st February Morning Shift - No. 12)
25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M $$\mathrm{AgNO_3}$$ solution when titrated using $$\mathrm{K_2CrO_4}$$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer).
(Given : $$\mathrm{K_f=2.0~K~kg~mol^{-1}}$$)
Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL$$^{-1}$$
Answer
3
Explanation
$25 \times M=20 \times 1$
$$ \begin{aligned} M & =\frac{20}{25}=\frac{4}{5}=0.8 \\\\ \Delta T_{f} & =(\mathrm{i})\left(\mathrm{K}_{\mathrm{f}}\right)(\mathrm{m}) \\\\ & =(2)(2)\left(\frac{4}{5}\right)=\frac{16}{5}=3.2 \end{aligned} $$
$$ \begin{aligned} M & =\frac{20}{25}=\frac{4}{5}=0.8 \\\\ \Delta T_{f} & =(\mathrm{i})\left(\mathrm{K}_{\mathrm{f}}\right)(\mathrm{m}) \\\\ & =(2)(2)\left(\frac{4}{5}\right)=\frac{16}{5}=3.2 \end{aligned} $$
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