The solution containing $18 \mathrm{~g} \mathrm{dm}^{-3}$ glucose (molar mass 180) in water and another…
The solution containing $18 \mathrm{~g} \mathrm{dm}^{-3}$ glucose (molar mass 180) in water and another containing $6 \mathrm{~g} \mathrm{dm}^{-3}$ of solute A in water boils at same temperature. What is molar mass of A ?
$54 \mathrm{~g} \mathrm{~mol}^{-1}$
$90 \mathrm{~g} \mathrm{~mol}^{-1}$
$120 \mathrm{~g} \mathrm{~mol}^{-1}$
$60 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution
As boiling points of both the solutions are same, they must have same molality.
$\begin{aligned}
\therefore \quad & \mathrm{m}_{\text {glucose }}=\mathrm{m}_{\mathrm{A}} \\
& \frac{\mathrm{~W}_{\text {glucose }}}{\mathrm{M}_{\text {glucose }} \times \mathrm{W}_{\text {solvent }}}=\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{A}} \times \mathrm{W}_{\text {solvent }}} \\
& \frac{18}{180 \times \mathrm{W}_{\text {water }}}=\frac{6}{\mathrm{M}_{\mathrm{A}} \times \mathrm{W}_{\text {water }}}
\end{aligned}$ Considering weight of water is same in both solutions,
$\therefore \quad M_A=60 \mathrm{~g} \mathrm{~mol}^{-1}$