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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:

A
0.200
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B
0.786
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C
0.549
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D
0.478
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Solution

The correct option is D 0.478
nC5H12nC6H14=14
xC5H12=15
and
xC6H14=45
poC5H12=440 mm Hg, poC6H14=120 mm Hg
Ptotal=poC5H12xC5H12+poC6H14xC6H14
=440×15+120×45=88+96=184 mm Hg
By Raoult's law,
pC5H12=poC5H12xC5H12...(i)
xC5H12- mole fraction of pentane in solution
By Dalton's law
PC5H12=xC5H12Ptotal...(ii)
xC5H12-Mole fraction of pentane above the solution
From (i) and (ii)
pC5H12=440×15=88 mm Hg
88=xC5H12×184
xC5H12=88184=0.478

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