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

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

JIPMERJIPMER 2007Solutions

Solution:

Total vapour pressure of mixture
$=$ (Mole fraction of pentane $\times$ VP of pentane) $+$ (Mole fraction of hexane $\times$ VP of hexane)
= VP of pentane in mixture $+$ VP of hexane in mixture.
$=\left(\frac{1}{5} \times 440+\frac{4}{5} \times 120\right)=184 \,mm$
$\because$ VP of Pentane in mixture.
$=$ VP of mixture $\times$ mole fraction of pentane in vapour phase
$88=184 \times$ mole fraction of pentane in vapour phase
$\therefore $ Mole fraction of pentane in vapour phase
$=\frac{88}{184}=0.478$