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Question
Heptane and octane form ideal solution. At 373 K, the vapour pressures of the pure liquids are 106 kPa and 46 kPa, respectively. What will be the vapour pressure, in bar, of a mixture of 30.0 g of heptane and 34.2 g of octane?
Options
A.76 bar
B.152 bar
C.1.52 bar
D.0.76 bar
Solution
$n_{\text{neptane}},C_{7}H_{16} = \frac{30}{0.3} = 0.3 = n_{1}$
$n_{\text{octane}},C_{8}H_{18} = \frac{34.2}{114} = 0.3 = n_{2}$
Now, $P = X_{1}.P_{1}^{o} + X_{2}.P_{2}^{o} = \frac{0.3}{0.6} \times 106 + \frac{0.3}{0.6} \times 46 = 76\text{ KPa = 76 bar}$
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