Question

The freezing point $\left(in{}^{\circ }C\right)$ of solution containing 0.1 g of $K_3\left[Fe\right(CN){]}_6$ (mol.wt.392) in 100 g of water($K_f$=1.86K kg $mo{l}^{-1}$) 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 $-2.3\times {10}^{-2}$

Van’t Hoff factor (i) $=4\{3K^{+}+[Fe\left(CN{)}_6{]}^{3-}\right\}$
$Molality=\frac{0.1}{329}\times \frac{1000}{100}=\frac{1}{329}$
$\Rightarrow -\Delta T_f=i{K}_f.m$
$=4\times 1.86\times \frac{1}{329}=2.3\times {10}^{-2}$
$\Rightarrow T_f=-2.3\times {10}^{{-2}^{\circ }}C$
(As freezing point of water is $0^{\circ }C$)