Question

If the boiling point elevation constant and the boiling point of benzene are $2.16{\text{K kg mol}}^{-1}$ and ${81.0}^{\circ }\text{C}$ respectively, calculate the boiling point of a solution formed when $5\text{g}$ of $C_{14}{H}_{12}$ is dissolved in $15\text{g}$ of benzene.

• A. ${90}^{\circ }\text{C}$
• B. ${75}^{\circ }\text{C}$
• C. ${81}^{\circ }\text{C}$
• D. ${85}^{\circ }\text{C}$

Answer 1

The correct option is D ${85}^{\circ }\text{C}$

$\Delta T_b=k_b\times m$
$molality=\frac{\text{number of moles of solute}}{\text{weight of solvent}}\times 1000$
Molar mass of $C_{14}{H}_{12}$ = 180 g/mol
$molality=\frac{5/180}{15}\times 1000=1.85m$
$\Delta T_b=k_b\times m$
$=2.16\times 1.85\approx 4K$
Boiling point of the solution = ${85}^{o}C$