Question

0.01 M solution of $KCl$ and $CaC{l}_2$ are prepared in water. The freezing point of $KCl$ is found to be $-2^{o}C$. What is the freezing point of $CaC{l}_2$ to be completely ionised?

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

Answer 1

The correct option is A $-3^{o}C$

$KCl⟶K^{+}+C{l}^{-}$

$i$ for $KCl=2,$

$CaC{l}_2⟶C{a}^{2+}+2C{l}^{-}$

$i$ for $CaC{l}_2=3$

$△T_l\propto i$

$\frac{△T_l\left(KCl\right)}{△T_l\left(CaC{l}_2\right)}=\frac{2}{3}$

$△T_l\left(CaC{l}_2\right)=\frac{3}{2}\times 2=3^{o}C$

Freezing point of $CaC{l}_2=0-3^{o}C=-3^{0}C$

Hence, option A is correct.