InterviewSolution
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InterviewSolution
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The molar volume of liquid benzene (density `0.877 g mL^(-1))` increases by a factor of `2750` as it vapourises at `20^(@)C` and that of liquid toluene(density `0.867 g mL)` increases by a factor of `7720`at `20^(@)C`. A solution of benzene and toluene at `20^(@)C` has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution. |
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Answer» Correct Answer - `0.732` `V_(B) = (78)/(0.877) xx 2750 mL = 244.583 L` `V_(T) = (92)/(0.867) xx 7720 mL = 819.192 L` `P_(B) = (1 xx 0.0821 xx293 xx760)/(244.583) = 74.74` torr `P_(T) = (1 xx 0.0821 xx 293 xx 760)/(819.192) =22.317` `46 = 74.74 X_(B) +23.317 (1-X_(B))` `52.423 X_(B) = 23.683` `X_(B) = 0.451` `Y_(B) = (1 xx 0.0821 xx 293 xx 760)/(819.192) = 22.317` `46 = 74.74 X_(B) +23.317 (1-X_(B))` `54.423 X_(B) = 23.683` `X_(B) = 0.451` `Y_(B) = (P_(B)^(@)-X_(B))/(P_(T)) = (74.74xx 0.451)/(46) = 0.732` |
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