According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

Given: Mass of solute ($w_2$) = 8 g Mass of solvent ($w_1$) = 114 g Molar mass of solvent ($M_1$, n-octane) = 114 g mol⁻¹ Vapour pressure is reduced to 80%, so the relative lowering is ($100 - 80$)/100 = 0.20.

Formula: For dilute solutions, the approximation $\frac{p_1^0 - p_1}{p_1^0} \approx \frac{n_2}{n_1}$ is often used . $\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 \times M_1}{M_2 \times w_1}$

Calculation:

1. Calculate moles of solvent ($n_1$): $n_1 = 114 \text{ g} / 114 \text{ g mol}^{-1} = 1 \text{ mol}$.
2. Apply the formula: $0.20 = \frac{8 \text{ g} \times 114 \text{ g mol}^{-1}}{M_2 \times 114 \text{ g}}$ $0.20 = \frac{8}{M_2}$ $M_2 = \frac{8}{0.20} = 40 \text{ g mol}^{-1}$

(Note: The approximation is used here to match the options provided in the examination).