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 (molar mass= 100 g mol-1) and 35.0 g of octane (molar mass of octane= 114 g mol -1)will be

Amount of heptane, $n_{1} = \frac{m_{1}}{M_{1}} = \frac{25.0 g}{100 g {mol}^{- 1}} = 0.25 mol$

Amount of octane, $n_{2} = \frac{m_{2}}{M_{2}} = \frac{35.0 g}{114 g {mol}^{- 1}} = 0.307 mol$

Mole fraction of heptane, $x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{0.25}{0.25 + 0.307} = 0.45$

Mole fraction of octane, $x_{2} = 1 - x_{x} = 0.55$

According to Raoult's law $p = x_{1} p_{1}^{*} + x_{2} p_{2}^{*} = (0.45) (105 kPa) + (0.55) (45 kPa) = 72.0 kPa$

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