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The freezing point of 1 molal ${CaCl}_{2}$ solution is −3.62°C . The percentage ionization of ${CaCl}_{2}$ is

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94.6%

47.3%

69.73%

23.65 %

answer is C.

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Detailed Solution

$∆ T_{f} = K_{f} \times m$

$= 1.86 \times 1$

=1.86

$i = \frac{3.62}{1.86} = 1.94$

$α = \frac{i - 1}{n - 1} = \frac{1.94 - 1}{2} = 0.473$

=47.3%

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