Question

Solution is made by dissolving $0.518$ mol of nonelectrolyte solution in $887$ g of benzene. Calculate the freezing point. $T_f$ and boiling point, $T_b$ of the solution.

Answer 1

Molality of solution = $m-\frac{0.518}{887g}=0.58399$

Elevation in boiling point $\Delta T_b=K_{b}m$

$\Delta T_b=2.53\times 0.58399$

$={1.478}^{o}C$

$T_b=T_b^o+\Delta T_b$

$=80.1+1.478={81.578}^{o}C$

depression in freeing point $\Delta T_f=K_{f}m$

$=5.12\times 0.58399$

$=3^{o}C$

$T_t=T_f^o=\Delta T_f=5.5-3^{o}C={2.5}^{o}C$