Let the molar mass of solute \(=M \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\therefore\) Moles of solute present \(=\frac{30 g}{\mathrm{M} \mathrm{g} \mathrm{mol}^{-1}}=\frac{30}{M} \mathrm{~mol}\)
Moles of solvent present, \(\left(n_1\right)=\frac{90}{18}=5\) moles.
\(\therefore \quad \frac{P^{\circ}-P_s}{P^{\circ}}=\frac{n_2}{n_1+n_2}\)
\(\frac{P^{\circ}-2 \cdot 8}{P^{\circ}}=\frac{30 / M}{5+30 / M}\)
\(1-\frac{2 \cdot 8}{P^{\circ}}=\frac{30}{(5 M+30)}\)
\(1-\frac{30}{5 M+30}=\frac{2 \cdot 8}{P^{\circ}}\)
\(1-\frac{6}{M+6}=\frac{2.8}{P^{\circ}}\)
\(\frac{M+6-6}{M+6}=\frac{2 \cdot 8}{P^{\circ}}\)
\(\frac{M}{M+6}=\frac{2 \cdot 8}{P^{\circ}}\)
\(\frac{P^o}{2 \cdot 8}=1+\frac{6}{M}\) ...(i)
After adding \(18 g\) of water,
Moles of water becomes \(=\frac{90+18}{18}=\frac{108}{18}=6\) moles
\(\therefore \frac{P^{\circ}-P_s}{P^{\circ}}=\frac{30 / M}{6+30 / M}\)
New vapour pressure \(=2.9 \mathrm{kPa}\)
\(\begin{aligned}
&\frac{P^{\circ}-2 \cdot 9}{P^{\circ}}=\frac{30 M}{M(6 M+30)}=\frac{5}{M+5} \\
&1-\frac{2 \cdot 9}{P^{\circ}}=\frac{5}{M+5} \\
&1-\frac{5}{M+5}=\frac{2 \cdot 9}{P^{\circ}} \\
&\frac{M+5-5}{M+5}=\frac{2 \cdot 9}{\mathrm{P}^{\circ}} \\
&\frac{P^{\circ}}{2 \cdot 9}=\frac{M+5}{M} \Rightarrow=1+\frac{5}{M} \\
&\frac{P^{\circ}}{2 \cdot 9}=1+\frac{5}{M} ...(ii)
\end{aligned}\)
Dividing equation \((i)\) by \((\) ii \()\), we get,
\(\begin{aligned}
&\frac{2 \cdot 9}{2 \cdot 8}=\frac{1+6 / M}{1+5 / M} \\
&2 \cdot 9\left(1+\frac{5}{M}\right)=2 \cdot 8\left(1+\frac{6}{M}\right) \\
&2 \cdot 9+\frac{2 \cdot 9 \times 5}{M}=2 \cdot 8+\frac{2 \cdot 8 \times 6}{M} \\
&2 \cdot 9+\frac{14 \cdot 5}{M}=2 \cdot 8+\frac{16 \cdot 8}{M} \\
&0 \cdot 1=\frac{16 \cdot 8}{M}-\frac{14 \cdot 5}{M}=\frac{2 \cdot 3}{M} \\
&M=\frac{2 \cdot 3}{0 \cdot 1} \\
&M=23 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)
Putting \(\mathrm{M}=23\), in equation ( \(i\) ), we get,
\(\begin{aligned}
&\frac{P^{\circ}}{2 \cdot 8}=1+\frac{6}{23}=\frac{29}{23} \\
&P^{\circ}=\frac{29}{23} \times 2 \cdot 8=3.53 \mathrm{kPa} .
\end{aligned}\)