The vapour pressure of hexane $(C_{6} H_{14})$ and heptane $(C_{7} H_{16})$ at $50^{o} C$ are 408 Torr and 141 Torr, respectively. The composition of the vapour above a binary solution composition containing a mole fraction of 0.300 hexane is ( $Y_{6} =$ mol fraction of hexane and $Y_{7} =$ mol fraction of heptane).

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$Y_{6} = 0 .8, Y_{7} = 0 .2$

$Y_{6} = 0 .554, Y_{7} = 0 .446$

$Y_{6} = 0 .300, Y_{7} = 0 .700$

$Y_{6} = 0 .871, Y_{7} = 0 .129$

answer is B.

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Detailed Solution

$Y_{6} = \frac{P_{6}^{o} X_{6}}{P_{T}} = \frac{408 \times 0 .3}{P_{T}} . .. (1)$

$Y_{7} = \frac{P_{7}^{o} X_{7}}{P_{T}} = \frac{141 \times 0 .7}{P_{T}} . .. (2)$

Dividing equation (1) by (2), we get

$\frac{Y_{6}}{Y_{7}} = 1 .24 & Y_{6} + Y_{7} = 1$

$∴ Y_{7} = 0 .446 & Y_{6} = 0 .554$

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