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  4. A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be:

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Q. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at $20^{\circ}C$ are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in vapour phase would be:

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Solution:

$P=P^{o}_{C_5H_{12}}\times X_{C_5 H_{12}}+P_{C_6 H_{14}}^{o} \times X_{C_6 H_{14}}$;
$p=440\times\frac{1}{5}+120\times\frac{4}{5}=\frac{920}{5}=184$
$y_{c_5 H_{12}}=\frac{P_{C_5 H_{12}}}{P_{t}}$
Now, $Y_{C_5 H_{12}}=\frac{88}{184}=0.478$