A solution contains 1 mol of pentane p∗=450mmHg and 4 mol of hexane p∗=150mmHg, the mole fraction of pentane in vapour phase will be

A solution contains 1 mol of pentane p=450mmHg and 4 mol of hexane p=150mmHg, the mole fraction of pentane in vapour phase will be

  1. A

    0.454

  2. B

    0.429

  3. C

    0.641

  4. D

    0.75

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    Solution:

    In vapour phase, we will have

    ppentane ={1/(1+4)}(450mmHg)=90mmHg; ppentane ={4/(1+4)}(150mmHg)=120mmHgypentane =ppentane ppentane +phexane =9090+120=0.429

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