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Q.

25 mL of an aqueous solution of $KCl$ was found t require 20 mL of 1M ${AgNO}_{3}$ solution when titrated using $K_{2} {CrO}_{4}$ as indicator. Depression in freezing point of $KCl$ solution with 100% ionization will be :
$(K_{f} = \frac{10}{9} k / molal)$

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a

$\frac{80}{45}$

b

$\frac{160}{45}$

c

$\frac{20}{45}$

d

$\frac{40}{45}$

answer is B.

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Detailed Solution

$\begin{array}{l} 25 \times M = 20 \times 1 \\ \Rightarrow M = 0.8 \\ Δ T_{f} = i \cdot K_{f} \cdot m = 2 \times \frac{10}{9} \times 0.8 \\ = \frac{16}{9} or \frac{80}{45} \end{array}$

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25 mL of an aqueous solution of KCl was found t require 20 mL of 1M AgNO3 solution when titrated using K2CrO4 as indicator. Depression in freezing point of KClsolution with 100% ionization will be :Kf=109k/molal