On mixing, heptane and octane form an ideal solution. At $$373$$ K, the vapour pressures of the two liquid components (heptane and octane) are $$105$$ kPa and $$45$$ kPa respectively. Vapour pressure of the solution obtained by mixing $$25.0$$ g of heptane and $$35$$ g of octane will be (molar mass of heptane $$= 100$$ g mol$$^{-1}$$ and of octane $$= 114$$ g mol$$^{-1}$$).

For an ideal binary solution Raoult’s law states that the total vapour pressure at a given temperature is the mole-fraction-weighted sum of the pure-component vapour pressures:

$$P_{\text{total}} \;=\; x_{\text{hep}}\,P_{\text{hep}}^\ast \;+\; x_{\text{oct}}\,P_{\text{oct}}^\ast$$

Step 1 - moles of each component
Heptane: $$n_{\text{hep}} \;=\;\dfrac{25.0\ \text{g}}{100\ \text{g mol}^{-1}} \;=\;0.25\ \text{mol}$$
Octane: $$n_{\text{oct}} \;=\;\dfrac{35.0\ \text{g}}{114\ \text{g mol}^{-1}} \;=\;0.307\ \text{mol}$$

Step 2 - mole fractions
Total moles: $$n_{\text{tot}} = 0.25 + 0.307 = 0.557\ \text{mol}$$
$$x_{\text{hep}} = \dfrac{0.25}{0.557} = 0.448$$
$$x_{\text{oct}} = \dfrac{0.307}{0.557} = 0.552$$

Step 3 - apply Raoult’s law (given $$P_{\text{hep}}^\ast = 105\ \text{kPa},\; P_{\text{oct}}^\ast = 45\ \text{kPa}$$)
$$\begin{aligned} P_{\text{total}} &= 0.448\,(105\ \text{kPa}) \;+\; 0.552\,(45\ \text{kPa})\\ &= 47.04\ \text{kPa} + 24.84\ \text{kPa}\\ &= 71.88\ \text{kPa}\,\approx\,72.0\ \text{kPa} \end{aligned}$$

Hence, the vapour pressure of the solution is $$72.0$$ kPa.

Option A which is: $$72.0$$ kPa