Question

The freezing point ( in ${ }^{\circ} C$ ) of a solution containing $0.1\, g$ of $K_3 Fe ( CN )_6$ (Mol. wt. $329$) in $100\, g$ of water $K_f=1.86\, K\,kg\,mol ^{-1}$ 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

Answer: (A)

$\left(\Delta T_F\right)_{\text {solution }}=\left(\Delta T_F\right)_{\text {solvent }}-\left(\Delta T_F\right)_{\text {solvent with solute }}$
$\left(\Delta T_F\right)_{\text {Solute }}=\left(\right.$ molality $\left.\times K_F \times i\right)$
$=\frac{0.1}{329 \times 100} \times 1000 \times 4 \times 1.86$
$=0.022^{\circ} C$
$\left(\Delta T_F\right)_{\text {solvent }}=0^{\circ} C$
So $\left(\Delta T_F\right)_{\text {Solution }}=0-0.02232$
$\cong-2.3 \times 10^{-2}$