According to Raoult's law, for an ideal solution of volatile liquids, the total vapour pressure ($p_{total}$) is the sum of the partial vapour pressures of the components . $p_{total} = p_1^0 x_1 + p_2^0 x_2$

Let benzene be component 1 and octane be component 2. Given molar ratio $n_1 : n_2 = 3 : 2$. Mole fraction of benzene, $x_1 = \frac{n_1}{n_1 + n_2} = \frac{3}{3 + 2} = \frac{3}{5} = 0.6$ Mole fraction of octane, $x_2 = \frac{n_2}{n_1 + n_2} = \frac{2}{3 + 2} = \frac{2}{5} = 0.4$

Given vapour pressures of pure components: $p_1^0 = 280 \text{ mm Hg}$ $p_2^0 = 420 \text{ mm Hg}$

Total vapour pressure, $p_{total} = (280 \times 0.6) + (420 \times 0.4)$ $p_{total} = 168 + 168 = 336 \text{ mm Hg}$.