Question

If $30g$ of a solute of molecular weight $154gmo{l}^{-1}$is dissolved in $250g$ of benzene, what will be the boiling point of the resulting solution under atmospheric pressure?
The molal boiling-point elevation constant for benzene is $2.61Kkgmo{l}^{-1}$ and the boiling point of pure benzene is ${80.1}^{o}C.$

• A. ${84.1}^{\circ }C$
• B. ${80.1}^{\circ }C$
• C. ${82.1}^{\circ }C$
• D. ${85.1}^{\circ }C$

Answer 1

The correct option is C ${82.1}^{\circ }C$

Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.
Molality $=\frac{△T_b}{K_b}=\frac{△T_b}{2.61}$
Given,
Mass of the solute = $30g$
No. of moles of solute $=\frac{30}{154}$
$\text{Molality (m)}=\frac{n_{\text{solute}}\times 1000}{W_{\text{solvent}}\text{in gram}}$
$\therefore$
Molality $=\frac{30}{154}\times \frac{1000}{250}mol/kg$
Thus, $\frac{△T_b}{2.61}=\frac{30}{154}\times \frac{1000}{250}$
$△T_b=2K$
Since, the boiling point of pure solvent is ${80.1}^{\circ }C$
The boiling of the resulting solution is $80.1+2={82.1}^{\circ }C.$