Question

If the boiling point of an aqueous solution containing a non-volatile solute is ${100.15}^{\circ }C$. What is its freezing point? Given latent heat of fusion and vapourization of water $80cal{g}^{-1}$ and $540cal{g}^{-1}$, respectively.

• A. $-{0.542}^{\circ }C$
• B. $-{1.542}^{\circ }C$
• C. $-{2.542}^{\circ }C$
• D. $-{1.542}^{\circ }C$

Answer 1

The correct option is A $-{0.542}^{\circ }C$

For a given aqueous solution,
$\Delta T_f=K_f\times m....\left(i\right)$
$\Delta T_b=K_b\times m....\left(ii\right)$
$K_f=\frac{R{T}_f^2}{1000l_f}....\left(iii\right)$
$K_b=\frac{R{T}_v^2}{1000l_v}....\left(iv\right)$
Dividing Eq. (iii) by Eq. (iv),
$\frac{k_f}{K_b}=\frac{T_f^2\times l_V}{T_b^2\times l_f}$
$\therefore \frac{\Delta T_f}{\Delta T_b}=\frac{T_f^2\times l_v}{T_b^2\times l_f}$
$T_f=0+273=273K$,
$T_b=100+273=373K$
$l_f=80cal{g}^{-1},lv=540cal{g}^{-1}$
$\therefore \frac{\Delta T_f}{0.15}=\frac{273\times 273\times 540}{373\times 373\times 80}$
$\Delta T_f=\frac{\Delta T_f}{0.15}=\frac{273\times 273\times 540}{373\times 373\times 80}\times 0.15=0.15=0.542$
$\therefore T_F=0-0.542=-{0.542}^{\circ }C$