Q75 / 160 · JIPMER 2007 · Chemistry

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

  • A
    $0.549$
  • B
    $0.200$
  • C
    $0.786$
  • Correct answer
    D
    $0.478$

Solution

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 $=\frac{88}{184}=0.478$

Hence, the correct answer is option: 'D'.

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