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$0.01 M$ solution of $KCl and {BaCl}_{2}$ are prepared in water.
The freezing points of $KCl$ is found to be -2° C. What is
the freezing point of ${BaCl}_{2}$ to be completely ionized

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$- 3 ° C$

$+ 3 ° C$

$- 2 ° C$

$- 4 ° C$

answer is A.

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Detailed Solution

i for $KCl = 2, i for {BaCl}_{2} = 3 ∵ Δ T_{f} \propto i$

$\frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{2}{3} \Rightarrow Δ T_{f} ({BaCl}_{2}) = \frac{3}{2} \times 2 = 3^{\circ} C$

$∴$ Freezing point of ${BaCl}_{2} = - 3^{\circ} C .$

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