Question

A solution of $2.5g$ of a non - volatile solid in $100g$ benzene is boiled at $0.42K$ higher than the boiling point of pure benzene. Calculate the molar mass of the substance. Molal elevation constant of benzene is $2.67Kkgmo{l}^{-1}$

• A. $192.9g/mol$
• B. $140.5g/mol$
• C. $201.5g/mol$
• D. $158.9g/mol$

Answer 1

The correct option is D $158.9g/mol$

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.

$\text{Molality (m)}=\frac{n_{\text{solute}}\times 1000}{W_{\text{solvent}}\text{in gram}}$
$\text{Molality (m)}=\frac{w_{\text{solute}}\times 1000}{M_{\text{solute}}.W_{\text{solvent}}\text{in gram}}$
where,
$w_{\text{solute}}$ is mass of solute
$W_{\text{solvent}}$ is mass of solvent
$M_{\text{solute}}$ is molar mass of solute
$\therefore$
$△T_b=\frac{1000K_b\times w}{W\times M}M=\frac{1000K_b\times w}{W\times △T_b}\text{Given,}{K}_b=2.67Kkgmo{l}^{-1}w=2.5g,W=100g,△T_b=0.42KM=\frac{1000\times 2.67\times 2.5}{100\times 0.42}$
The molar mass of substance is $158.9g/mol$