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JIPMER 2007
Exam
JIPMER
Year
2007
Subject
Chemistry
Chapter
Solutions

JIPMER 2007 · Chemistry · Solutions

A solution has a 1: 4 mole ratio of pentane to hexane. The vapour...

A solution has a 1:41: 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20∘C20^{\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

  1. A0.5490.549
  2. B0.2000.200
  3. C0.7860.786
  4. D0.4780.478

Correct answer

D. 0.4780.478

Step-by-step solution

Total vapour pressure of mixture = (Mole fraction of pentane ×VP\times \mathrm{VP} of pentane )) + (Mole fraction of hexane ×VP\times \mathrm{VP} of hexane) =VP=\mathrm{VP} of pentane in mixture +VP of hexane in mixture =(15×440+45×120)=184 mm=\left(\frac{1}{5} \times 440+\frac{4}{5} \times 120\right)=184 \mathrm{~mm} ∵\because VP of Pentane in mixture. =VP=\mathrm{VP} of mixture ×\times mole fraction of pentane in vapour phase 88=184×88=184 \times mole fraction of pentane in vapour phase ∴\therefore Mole fraction of pentane in vapour phase =88184=0.478=\frac{88}{184}=0.478
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