Question

If boiling point of an aqueous solution is ${100.1}^{\circ }C$, calculate its freezing point.
Given, enthalpy of fusion and vaporisation of water are $80cal{g}^{-1}$ and $540cal{g}^{-1}$ respectively

(Take $\frac{273}{373}\approx 0.7$)

• A. $+{0.33}^{\circ }C$
• B. $-{0.33}^{\circ }C$
• C. $-{1.55}^{\circ }C$
• D. $+{1.55}^{\circ }C$

Answer 1

The correct option is B $-{0.33}^{\circ }C$

Given $\Delta H_{fus}=80cal{g}^{-1}$
$\Delta H_{vap.}=540cal{g}^{-1}$
Since,
$\Delta T_b=K_b\times m$
and $\Delta T_f=K_f\times m$
Here m is molality
$K_b\&K_f\text{are ebullioscopic constant and cryoscopic constant respectively}$

Also $K=\frac{R{T}^2}{1000\times \Delta H}$

$\therefore \frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}$

$\Rightarrow \frac{\Delta T_b}{\Delta T_f}=\frac{R{T}_b^2}{1000\times \Delta H_{vap.}}\times \frac{1000\times \Delta H_{fus.}}{R{T}_f^2}$

$\frac{\Delta T_b}{\Delta T_f}=\frac{T_b^2\times \Delta H_{fus}}{T_f^2\times \Delta H_{vap.}}$

$\frac{0.1}{\Delta T_f}=\frac{373\times 373\times 80}{273\times 273\times 540}$

$\frac{\Delta T_f}{0.1}=\frac{(0.7{)}^2\times 540}{80}$
$\Delta T_f=0.33T_f^{\circ }-T_f=0.330-T_f=0.33T_f=-{0.33}^{\circ }C$