Question

$0.01 \,M$ solution of $KCl$ and $CaCl _{2}$ are prepared in water. The freezing point of $KCl$ is found to be $-2^{\circ} C .$ What is the freezing point of $CaCl _{2}$ to be completely ionised?

• A. $-3^{\circ} C$
• B. $+3^{\circ} C$
• C. $-2^{\circ} C$
• D. $-4^{\circ} C$

Answer 1

Answer: (A)

$i$ for $K C l=2i$, for for $C a C l_{2}=3$
$\Delta T_{f} \propto i$
$\frac{\Delta T_{f}(K C l)}{\Delta T_{f}\left(C a C l_{2}\right)}=\frac{2}{3}$
$\Delta T_{f}\left(CaCl_{2}\right)=\frac{3}{2} \times 2=3^{\circ} C$
Freezing point of $CaCl_{2}=3^{\circ} C$