Question

Find the boiling point of a solution of 5.00 g of naphthalene $\left(C_{10}{H}_8\right)$ in 100 g of benzene. $K_b$ of benzene = 2.53${}^o$C/m; the normal boiling point of benzene = 80${}^o$C.

• A. 81${}^o$C
• B. 85${}^o$C
• C. 0.99${}^o$C
• D. 79${}^o$C

Answer 1

The correct option is A 81${}^o$C

Molar mass of naphthalene is 128 g/mol.

Number of moles of naphthalene are $\frac{5.00g}{128g/mol}=0.039$ moles.

Mass of benzene $=100$ g $=0.100$ kg.

Molality, $m=\frac{0.039mol}{0.100kg}=0.39mol/kg$

The elevation in the boiling point $\Delta T_b=K_{b}m=2.53\times 0.39={0.987}^{0}C\simeq 1^{o}C$

The boiling point of pure benzene is $80{}^{o}C$

The boiling point of solution $=80{}^{o}C+1{}^{o}C=81{}^{o}C$