$A$ solution containing $30 \,g$ of non-volatile solute in $90 \,g$ of water has a vapour pressure of $2.8 \,kPa$ at $298 \,K$. Further,$18 \,g$ of water is added to the solution and the new vapour pressure becomes $2.9 \,kPa$ at $298 \,K$. Calculate:
$i$. Molar mass of the solute
$ii$. Vapour pressure of water at $298 \,K$

$(i)$ Let the molar mass of the solute be $M \,g \,mol^{-1}$.
Initial state:
$n_{1} = \frac{90 \,g}{18 \,g \,mol^{-1}} = 5 \,mol$ (moles of water)
$n_{2} = \frac{30 \,g}{M \,g \,mol^{-1}} = \frac{30}{M} \,mol$ (moles of solute)
$p_{1} = 2.8 \,kPa$
Using Raoult's Law: $\frac{p_{1}^{o} - p_{1}}{p_{1}^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$1 - \frac{2.8}{p_{1}^{o}} = \frac{30/M}{5 + 30/M} = \frac{30}{5M + 30}$
$\frac{2.8}{p_{1}^{o}} = 1 - \frac{30}{5M + 30} = \frac{5M}{5M + 30}$
$\frac{p_{1}^{o}}{2.8} = \frac{5M + 30}{5M} \quad \dots (i)$
After adding $18 \,g$ of water:
$n_{1}' = \frac{90 + 18}{18} = 6 \,mol$
$p_{1}' = 2.9 \,kPa$
$\frac{p_{1}^{o} - 2.9}{p_{1}^{o}} = \frac{30/M}{6 + 30/M} = \frac{30}{6M + 30}$
$\frac{p_{1}^{o}}{2.9} = \frac{6M + 30}{6M} \quad \dots (ii)$
Dividing $(i)$ by $(ii)$:
$\frac{2.9}{2.8} = \frac{(5M + 30)/5M}{(6M + 30)/6M} = \frac{5M + 30}{5} \times \frac{6}{6M + 30} = \frac{6(5M + 30)}{5(6M + 30)}$
$2.9 \times 5(6M + 30) = 2.8 \times 6(5M + 30)$
$14.5(6M + 30) = 16.8(5M + 30)$
$87M + 435 = 84M + 504$
$3M = 69 \Rightarrow M = 23 \,g \,mol^{-1}$.
$(ii)$ Substituting $M = 23$ in $(i)$:
$\frac{p_{1}^{o}}{2.8} = \frac{5(23) + 30}{5(23)} = \frac{115 + 30}{115} = \frac{145}{115} \approx 1.2608$
$p_{1}^{o} = 2.8 \times 1.2608 = 3.53 \,kPa$.