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

Now, the no. of moles of solvent (water), n₁ = \(\frac{90g}{18\;g \;mol^{-1}}\) = 5 mol

And, the no. of moles of solute, n₂ = \(\frac{30g}{M \;mol^{-1}}\) = \(\frac{30}{M}\) mol

P₁ = 2.8 kPa

Applying the relation,

\(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{n_2}{n_1 + n_2}\)

⇒ \(\frac{p_1^0 - 2.8}{p_1^0}\) = \(\frac{\frac{30}{M}}{5 + \frac{30}{M}}\)

⇒ 1 - \(\frac{2.8}{p_1^0}\) = \(\frac{\frac{30}{M}}{ \frac{5M + 30}{M}}\)

⇒ 1 - \(\frac{2.8}{p_1^0}\) = \( \frac{30}{5M + 30}\)

⇒ \(\frac{2.8}{p_1^0}\) = \( 1 -\frac{30}{5M + 30}\)

⇒ \(\frac{2.8}{p_1^0}\) = \( \frac{5M + 30 - 30}{5M + 30}\)

⇒ \(\frac{2.8}{p_1^0}\) = \(\frac{5M}{5M + 30}\)

⇒ \(\frac{p_1^0}{2.8}\) = \(\frac{5M + 30}{5M}\)  ......(i)

After the addition of 18 g of water

n₁ = \(\frac{90+18g}{18}\) = 6 mol

p₁ = 2.9 kPa

Again, applying the relation,

\(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{n_2}{n_1 + n_2}\)

⇒ \(\frac{p_1^0 - 2.9}{p_1^0}\) = \(\frac{\frac{30}{M}}{6 + \frac{30}{M}}\)

⇒ 1 - \(\frac{2.9}{p_1^0}\) = \(\frac{\frac{30}{M}}{ \frac{6M + 30}{M}}\)

⇒ 1 - \(\frac{2.9}{p_1^0}\) = \( \frac{30}{6M + 30}\)

⇒ \(\frac{2.9}{p_1^0}\) = \( 1 -\frac{30}{6M + 30}\)

⇒ \(\frac{2.9}{p_1^0}\) = \( \frac{6M + 30 - 30}{6M + 30}\)

⇒ \(\frac{2.9}{p_1^0}\) = \(\frac{6M}{6M + 30}\)

⇒ \(\frac{p_1^0}{2.9}\) = \(\frac{6M + 30}{6M}\)  ......(ii)

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

⇒ \(\frac{2.9}{2.8}\) = \(\frac{\frac{5M + 30}{5M}}{\frac{6M + 30}{6M}}\)

⇒\(\frac{2.9}{2.8}\) x \(\frac{6M + 30}{6}\) = \(\frac{5M + 30}{5}\)

⇒ 2.9 x 5 x (6M + 30) = 2.8 x 6 x (5M + 30)

⇒ 87M + 435 = 84 M + 504

⇒ 3 M = 69

⇒ M = 23 u

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

(ii) Putting the value of 'M' in equation (i), we have

\(\frac{p_1^0}{2.8}\) = \(\frac{5 \times 23 + 30}{5 \times 23}\)

⇒ \(\frac{p_1^0}{2.8}\) = \(\frac{145}{115}\)

⇒ \(p_1^0\) = 3.53

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