Question

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

• A. ${0.361}^{\circ }C$
• B. $-{0.361}^{\circ }C$
• C. $-{3.61}^{\circ }C$
• D. None of these

Answer 1

Answer: (B)

$-{0.361}^{\circ }C$

(b) Given $\Delta H_{fus}=80cal{g}^{-1}$
$\Delta H_{vsp.}=540cal{g}^{-1}$
We know, $\Delta T_b=K_b\times m$
and $\Delta T_f=K_f\times m$;
Also $K=\frac{R{T}^2}{1000\times \Delta H}$
$\therefore \frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}$
$\Rightarrow \frac{R{T}_b^2}{1000\times \Delta H_{vap.}}\times \frac{1000\times \Delta H_{fus.}}{R{T}_f^2}$
$\Rightarrow \frac{\Delta T_b}{\Delta T_f}=\frac{T_b^2\times \Delta H_{fus}}{T_f^2\times \Delta H_{vap.}}$
$\frac{0.1}{\Delta T_f}=\frac{373\times 373\times 80}{273\times 273\times 540}$
$\because \Delta T_f=0.361$ so,
$T_f=-{0.361}^{\circ }C$