The boiling point of benzene is 353 K . ( $K_{b}$ for benzene is $2.53 K kg mol^{-1}$ ). When 1.80 g

Question

The boiling point of benzene is 353 K . ( $K_{b}$ for benzene is $2.53 K kg mol^{-1}$ ). When 1.80 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point is raised to 353.87 K . Molar mass of the solute is

Professor PC Thomas Classes · Answer

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}=\frac{\mathrm{K}_{\mathrm{b}} \mathrm{w}_{2} 	imes 1000}{\mathrm{M}_{2} \mathrm{w}_{1}}$
$$
\begin{aligned}
& \mathrm{M}_{2}=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{~W}_{2}}{\Delta \mathrm{~T}_{\mathrm{b}} \mathrm{w}_{1}} \
& \mathrm{M}_{2}=\frac{1000 	imes 2.53 	imes 1.80}{0.87 	imes 90}=58.16 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$