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Dalton's Law of Partial Pressures

A solution ha...

Question

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

0.200

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0.786

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0.549

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0.478

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Solution

The correct option is D 0.478
$\frac{n_{C_{5} H_{12}}}{n_{C_{6} H_{14}}} = \frac{1}{4}$
$\Rightarrow x_{C_{5} H_{12}} = \frac{1}{5}$
and
$x_{C_{6} H_{14}} = \frac{4}{5}$
$p^{o} C_{5} H_{12} = 440 m m H g, p^{o} C_{6} H_{14} = 120 m m H g$
$P_{t o t a l} = p_{C_{5} H_{12}}^{o} x C_{5} H_{12} + p_{C_{6} H_{14}}^{o} x C_{6} H_{14}$
$= 440 \times \frac{1}{5} + 120 \times \frac{4}{5} = 88 + 96 = 184 m m H g$
By Raoult's law,
$p_{C_{5} H_{12}} = p_{C_{5} H_{12}}^{o} x_{C_{5} H_{12}} . . . (i)$
$x_{C_{5} H_{12}}$ - mole fraction of pentane in solution
By Dalton's law
$P_{C_{5} H_{12}} = x_{C_{5} H_{12}}^{'} P_{t o t a l} . . . (i i)$
$x_{C_{5} H_{12}}^{'}$ -Mole fraction of pentane above the solution
From (i) and (ii)
$p_{C_{5} H_{12}} = 440 \times \frac{1}{5} = 88 m m H g$
$\Rightarrow 88 = x_{C_{5} H_{12}}^{'} \times 184$
$x_{C_{5} H_{12}}^{'} = \frac{88}{184} = 0.478$

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Similar questions

Q. A solution has a

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A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°Care 440mmHg for pentane and 120mm Hg for hexane. The mole fraction of pentane in the vapour phase would be:

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

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