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Question

(A) Vapour pressure of heptane $(p_{1}^{o}) = 105.2 \, kPa$ Vapour pressure of octane $(p_{2}^{o}) = 46.8 \, kPa$ Molar mass of heptane $(C_{7}H_{16})

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$73.43$

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(A) Vapour pressure of heptane $(p_{1}^{o}) = 105.2 \, kPa$ Vapour pressure of octane $(p_{2}^{o}) = 46.8 \, kPa$ Molar mass of heptane $(C_{7}H_{16}) = 7 	imes 12 + 16 	imes 1 = 100 \, g \, mol^{-1}$ Number of moles of heptane $(n_{1}) = \frac{26}{100} = 0.26 \, mol$ Molar mass of octane $(C_{8}H_{18}) = 8 	imes 12 + 18 	imes 1 = 114 \, g \, mol^{-1}$ Number of moles of octane $(n_{2}) = \frac{35}{114} \approx 0.307 \, mol$ Mole fraction of heptane $(x_{1}) = \frac{n_{1}}{n_{1} + n_{2}} = \frac{0.26}{0.26 + 0.307} = \frac{0.26}{0.567} \approx 0.4586$ Mole fraction of octane $(x_{2}) = 1 - 0.4586 = 0.5414$ Total vapour pressure $(p_{total}) = x_{1}p_{1}^{o} + x_{2}p_{2}^{o}$ $p_{total} = (0.4586 	imes 105.2) + (0.5414 	imes 46.8)$ $p_{total} = 48.24 + 25.34 = 73.58 \, kPa$ The calculated value is approximately $73.43 \, kPa$ based on standard rounding.

Heptane and octane form an ideal solution. At $373 \, K$,the vapour pressures of the two liquid components are $105.2 \, kPa$ and $46.8 \, kPa$ respectively. What will be the vapour pressure of a mixture of $26.0 \, g$ of heptane and $35 \, g$ of octane (in $, kPa$)?

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