(i) Let, the molar mass of the solute be M g mol⁻¹

Now, the no. of moles of solvent (water), ${\mathrm{n}}_1=\frac{90\mathrm{g}}{18 \mathrm{g} {\mathrm{mol}}^{-1}}=5 \mathrm{mol}$

And, the no. of moles of solute, ${\mathrm{n}}_2=\frac{30\mathrm{g}}{\mathrm{M} {\mathrm{mol}}^{-1}}=\frac{30}{\mathrm{M}}\mathrm{mol}$

${\mathrm{p}}_1=2.8 \mathrm{kPa}$

Applying the relation:

$$\begin{gathered} \frac{{\mathrm{p}}_1^0-{\mathrm{p}}_1}{{\mathrm{p}}_1^0}=\frac{n_2}{n_1+n_2} \\ \Rightarrow \frac{p_1^0-2.8}{p_1^0}=\frac{{\displaystyle \frac{20}{M}}}{5+{\displaystyle \frac{30}{M}}} \\ \Rightarrow 1-\frac{2.8}{p_1^0}=\frac{{\displaystyle \frac{30}{M}}}{{\displaystyle \frac{5+30}{M}}} \\ \Rightarrow 1-\frac{2.8}{p_1^0}=\frac{30}{5\mathrm{M}+30} \\ \Rightarrow \frac{2.8}{{\mathrm{p}}_1^0}=1-\frac{30}{5\mathrm{M}+30} \\ \Rightarrow \frac{2.8}{{\mathrm{p}}_1^0}=\frac{5\mathrm{M}+30-30}{5\mathrm{M}+30} \\ \Rightarrow \frac{2.8}{{\mathrm{p}}_1^0}=\frac{5\mathrm{M}}{5\mathrm{M}+30} \\ \Rightarrow \frac{{\mathrm{p}}_1^0}{2.8}=\frac{5\mathrm{M}}{5\mathrm{M}+30} ....\left(\mathrm{i}\right) \end{gathered}$$

After the addition of 18 g of water:

$$\begin{gathered} {\mathrm{n}}_1=\frac{90+18 \mathrm{g}}{18}=6 \mathrm{mol} \\ {\mathrm{p}}_1=2.9 \mathrm{kPa} \end{gathered}$$

Again, applying the relation:

$$\begin{gathered} \frac{{\mathrm{p}}_1^0-{\mathrm{p}}_1}{{\mathrm{p}}_1^0}-\frac{n_2}{n_1+n_2} \\ \Rightarrow \frac{p_1^0-2.9}{p_1^0}=\frac{{\displaystyle \frac{30}{M}}}{6+{\displaystyle \frac{30}{M}}} \\ \Rightarrow 1-\frac{2.9}{p_1^0}=\frac{{\displaystyle \frac{30}{M}}}{{\displaystyle \frac{6M+30}{M}}} \\ \Rightarrow 1-\frac{2.9}{p_1^0}=\frac{{\displaystyle 30}}{{\displaystyle 6\mathrm{M}+30}} \\ \Rightarrow \frac{2.9}{p_1^0}=1-\frac{{\displaystyle 30}}{{\displaystyle 6\mathrm{M}+30}} \\ \Rightarrow \frac{2.9}{p_1^0}=\frac{{\displaystyle 6\mathrm{M}+30-30}}{{\displaystyle 6\mathrm{M}+30}} \\ \Rightarrow \frac{2.9}{p_1^0}=\frac{{\displaystyle 6\mathrm{M}}}{{\displaystyle 6\mathrm{M}+30}} \\ \Rightarrow \frac{{\mathrm{p}}_1^0}{2.9}=\frac{{\displaystyle 6\mathrm{M}+30}}{{\displaystyle 6\mathrm{M}}}.....\left(\mathrm{ii}\right) \end{gathered}$$

Dividing equation (i) by (ii), we have:

$$\begin{gathered} \frac{2.9}{2.8}=\frac{{\displaystyle \frac{5M+30}{5M}}}{{\displaystyle \frac{6M+30}{6M}}} \\ \Rightarrow \frac{2.9}{2.8}\times \frac{6M+30}{6}=\frac{5M+30}{5} \\ \Rightarrow 2.9\times 5\times (6M+30)=2.8\times 6\times (5M+30) \\ \Rightarrow 87M+435=84M+504 \\ \Rightarrow 3M=69 \\ \Rightarrow M=23 u \end{gathered}$$

Therefore, the molar mass of the solute is 23 g mol⁻¹.

(ii) Putting the value of ‘M’ in equation (i), we have:

$\frac{{\mathrm{p}}_1^0}{2.8}=\frac{5\times 23+30}{5\times 23}$
$\Rightarrow \frac{{\mathrm{p}}_1^0}{2.8}=\frac{145}{115}$
$\Rightarrow {\mathrm{p}}_1^0=3.53$$$\begin{gathered} \frac{{\mathrm{p}}_1^0}{2.8}=\frac{5\times 23+30}{5\times 23} \\ \Rightarrow \frac{p_1^0}{2.8}=\frac{145}{115} \\ \Rightarrow p_1^0=3.53 \end{gathered}$$

Hence, the vapour pressure of water at 298 K is 3.53 kPa.