Question

Equimolal aqueous solution of $NaCl$ and $BaC{l}_2$ are prepared. If the freezing point of $NaCl$ is $-2^{o}C$, the freezing point of $BaC{l}_2$ solution is expected to be:

• A. $-2^{o}C$
• B. $-3^{o}C$
• C. $-{1.5}^{o}C$
• D. $-{1.66}^{o}C$

Answer 1

Answer: (B)

$-3^{o}C$

$△T_f=i{K}_{f}m$

$\Rightarrow N{a}^{+}+C{l}^{-}$ $i=2$

$△T_f=2k_{f}m=-2°C$

$\Rightarrow BaC{l}_2\to B{a}^{2+}+2C{l}^{-}$ $i=3$

$△T_f=3kfm$

Now, $2K_{f}m=-2°C$

$3K_{f}m=x$

$\Rightarrow x=\frac{3\times -2}{2}=-3°C$

$\therefore △T_f=-3°C$

Hence, the answer is $-3°C.$