Question

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

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

Answer 1

Answer: (A)

$- 3^{\circ}C$

Answer 2

$i$ for $KCl =2, i$ for $BaCl _{2}=3$ $\because \Delta T_{f} \propto i$ $\frac{\Delta T_{f}( KCl )}{\Delta T_{f}\left( BaCl _{2}\right)}=\frac{2}{3}$ $\Delta T_{f}\left( BaCl _{2}\right)=\frac{3}{2} \times 2=3^{\circ} C$ $\therefore$ Freezing point of $KCl =-3^{\circ} C$