Search any question & find its solution
Question:
Answered & Verified by Expert
A solution has a $1: 4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20^{\circ} \mathrm{C}$ 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
Options:
Solution:
1717 Upvotes
Verified Answer
The correct answer is:
$0.478$
Total vapour pressure of mixture $=$ (Mole fraction of pentane $\times \mathrm{VP}$ of pentane $)$ + (Mole fraction of hexane $\times \mathrm{VP}$ of hexane)
$=\mathrm{VP}$ of pentane in mixture +VP of hexane in mixture
$=\left(\frac{1}{5} \times 440+\frac{4}{5} \times 120\right)=184 \mathrm{~mm}$
$\because$ VP of Pentane in mixture. $=\mathrm{VP}$ 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 $=\frac{88}{184}=0.478$
$=\mathrm{VP}$ of pentane in mixture +VP of hexane in mixture
$=\left(\frac{1}{5} \times 440+\frac{4}{5} \times 120\right)=184 \mathrm{~mm}$
$\because$ VP of Pentane in mixture. $=\mathrm{VP}$ 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 $=\frac{88}{184}=0.478$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.