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A solution has 1:4 mole ratio of pentane to bexane. The vapour pressure of the pure hydrocarbhonse at `20^(@)C` and 440 mm Hg is for pentane and 120 mm Hg for bexane: The mole fraction of pentane in the vapour phase will be :A. 0.2B. 0.549C. 0.786D. 0.478 |
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Answer» Correct Answer - c `(n_(C_(5)H_(12)))/(n_(C_(6)H_(14)))=1/4` `X_(C_(5)H_(12))=1/(1+4)=1/5` `X_(C_(6)H_(14))=4/(1+4)=4/5` `P_(C_(5)H_(12))^(@)=440mmHg,P_(C_(6)H_(14))^(@)=120 mm Hg` `P_(("total"))=P_(C_(5)H_(12))^(@)xxX_(C_(5)H_(12))+P_(C_(6)H_(14))^(@)xxX_(C_(6)H_(14))` `=440xx1//5+120xx4//5` `=88+96=184 mm Hg` Mole fraction `C_(5)H_(12)` in vapour phase `X_(C_(5)H_(12))=("Partoa V.P. of "C_(5)H_(12))/("Partoa V.P. of "C_(6)H_(14))=(88mm)/(184mm)=0.478` |
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