The boiling point of benzene is $353.23 \, K$. When $1.80 \, g$ of a non-volatile solute was dissolved in $90 \, g$ of benzene,the boiling point is raised to $354.11 \, K$. Calculate the molar mass of the solute. $K_{b}$ for benzene is $2.53 \, K \, kg \, mol^{-1}$.

(N/A) The elevation in boiling point $(\Delta T_{b})$ is given by: $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 354.11 \, K - 353.23 \, K = 0.88 \, K$.
The formula for molar mass $(M_{2})$ of the solute is: $M_{2} = \frac{K_{b} \times w_{2} \times 1000}{\Delta T_{b} \times w_{1}}$.
Substituting the given values: $M_{2} = \frac{2.53 \, K \, kg \, mol^{-1} \times 1.80 \, g \times 1000 \, g \, kg^{-1}}{0.88 \, K \times 90 \, g} = 58 \, g \, mol^{-1}$.
Thus,the molar mass of the solute is $58 \, g \, mol^{-1}$.