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
A) 0.549
B) 0.200
C) 0.786
D) 0.478
Correct Answer: D
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 x 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 \[=\frac{88}{184}=0.478\]
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