Question

The freezing point (in ${}^{o}C$) of a solution containing 0.1 g of $K_3\left[Fe\right(CN{)}_6]$ (molar mass 329) in 100 g of water $(K_f=1.86K.kg.mo{l}^{-1})$ is :

• A. $-2.3\times {10}^{-2}{}^{o}C$
• B. $-5.7\times {10}^{-2}{}^{o}C$
• C. $-5.7\times {10}^{-3}{}^{o}C$
• D. $-1.2\times {10}^{-2}{}^{o}C$

Answer 1

The correct option is A $-2.3\times {10}^{-2}{}^{o}C$

The freezing point of the solution will be $=-2.3\times {10}^{-2}{}^{o}C$
$K_3\left[Fe\right(CN{)}_6]\to 3K^{+}+Fe(CN{)}_6^{3-}$
Before dissociation 1 0 0
After dissociation 0 3 1

Total no. of particles furnished by $K_3\left[Fe\right(CN{)}_6]=n=4$

$\therefore$ van't Hoff factor, $i=4$

Now,

$\Delta T_f=\frac{1000\times K_f\times w}{m\times W}\times i$

$=\frac{1000\times 1.86\times 0.1\times 4}{329\times 100}$

$=2.3\times {10}^{-2}{}^{o}C$

$\therefore T_f^{\prime }=0-2.3\times {10}^{-2}$

$\therefore T_f^{\prime }=-2.3\times {10}^{-2}{}^{o}C$