At \(298 \mathrm{~K}\), the vapour pressure of a solution of \(7.5 \mathrm{~g}\) of non-volatile solute in…
At \(298 \mathrm{~K}\), the vapour pressure of a solution of \(7.5 \mathrm{~g}\) of non-volatile solute in \(90 \mathrm{~g}\) of water is \(2.8 \mathrm{kPa}\). If \(18 \mathrm{~g}\) of water is added to this solution the vapour pressure becomes \(2.81 \mathrm{kPa}\) at same temperature, the molar mass of solute in \(\mathrm{g} \mathrm{mol}^{-1}\) is
17.5
68.2
71.5
51.8
Solution
Key Idea Relation between relative lowering of vapour pressure and molecular mass of solute is given by
\(\frac{p^{\circ}-p}{p^{\circ}}=\frac{w_2 / M_2}{w_1 M_1}\)
Given,
Weight of non-volatile solute, \(w_2=7.5 \mathrm{~g}\)
Weight of water, \(w_1=90 \mathrm{~g}\)
Vapour pressure of solution \(=2.8 \mathrm{kPa}\)
In first case,
\(\begin{aligned}
\frac{p^{\circ}-2.8}{p^{\circ}} & =\frac{7.5 / M_2}{90 / 18} \\
\frac{p^{\circ}-2.8}{p^{\circ}} & =\frac{1.5}{M_2} \quad \ldots (i)
\end{aligned}\)
In second case,
\(\frac{p^{\circ}-2.81}{p^{\circ}}=\frac{7.5 / M_2}{108 / 18}=\frac{1.25}{M_2} \quad \ldots (ii)\)
On solving (i) and (ii) we get
\(M_2=71.5 \mathrm{~g} \mathrm{~mol}^{-1}\)