Question

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. $\left(K_{b}forbenzeneis2.53Kkgmo{l}^{-1}\right)$

• A. $58.0gmo{l}^{-1}$
• B. $68.0gmo{l}^{-1}$
• C. $78.0gmo{l}^{-1}$
• D. $48.0gmo{l}^{-1}$

Answer 1

The correct option is A

$58.0gmo{l}^{-1}$

Given values are:
$T_{benzene}^{\circ }=353.00K;K_b=2.53Kkgmo{l}^{-1}$
$T_{b\left(solution\right)}=354.00k$
$W_{solute}=1.80g$
$W_{solvent}=90g$
The elevation in boiling point,$\Delta T_b=T_{b\left(solution\right)}-T_b^{\circ }\left(solvent\right)$
$=354.11-353.23$
$=0.88K$
Molar mass of solute is given as
$M{w}_{solute}=\frac{K_b\times 1000\times W_{solvent}}{\Delta T_b\times W_{solvent}}$
$M{w}_{solute}=\frac{2.53\times 1000\times 1.80}{0.88\times 90}=58.0gmo{l}^{-1}$
Hence, the molar mass of solute is $58.0gmo{l}^{-1}$