Question

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

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

Answer 1

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

$K_3\left[Fe\right(CN{)}_6]\to 3K^{+}+[Fe(CN{)}_6{]}^{3-}$

$i=4$

Now,

$\Delta T_f=K_f\times i\times \frac{m}{M}\times \frac{1000}{W}=1.86\times 4\times \frac{0.1}{329}\times \frac{1000}{100}=2.3\times {10}^{-2}$.

$T_{f^{\prime }}=0-2.3\times {10}^{-2}=-2.3\times {10}^{-2o}C$