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CBSE Class 12-science Answered

the molar volume of liquid benzene(density=0.867gm/ml)increased by a factor of 2750  at 20 degree celsius and that of liquid toluene(density=0.867gm/ml)increases by a factor of 7720 at 20 degree celsius . a solution of benzene and toluene at 20 degree celsius has a vapour pressure of 46 torr . find the mole fraction of benzene in the vapour above the solution.
Asked by vt6722 | 26 Jul, 2015, 01:32: PM
answered-by-expert Expert Answer
begin mathsize 11px style Given colon Density space of space benzene space equals space 0.877 space straight g space ml to the power of – 1 end exponent space space Molecular space mass space of space benzene space left parenthesis straight C subscript 6 straight H subscript 6 right parenthesis space space equals space 6 space cross times space 12 space plus space 6 space cross times space 1 space equals space 78  therefore space Molar space volume space of space liquid space benzene space equals space straight M over straight rho space equals space fraction numerator 78 space straight g space mol to the power of negative 1 end exponent over denominator 0.877 space gmL to the power of negative 1 end exponent end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space fraction numerator 78 over denominator 0.877 end fraction space cross times space 1 over 1000 straight L space equals space 244.58 space straight L Density space of space toluene space equals space 0.867 space straight g space ml – 1 space space Molecular space mass space of space toluene space left parenthesis straight C 6 straight H 5 CH 3 right parenthesis equals space 6 space cross times space 12 space plus space 5 space cross times space 1 space plus space 1 space cross times space 12 space plus space 3 space cross times space 1 space equals space 92 space therefore space Molar space volume space of space toluene space in space liquid space form space equals space fraction numerator 92 over denominator 0.867 end fraction ml space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 92 over denominator 0.867 end fraction space cross times space 1 over 1000 straight L And space molar space volume space of space toluene space in space vapour space phase space equals space fraction numerator 92 over denominator 0.867 end fraction space cross times space 7720 over 1000 straight L space equals space 819.19 space straight L Using space the space ideal space gas space equation comma space space PV space equals space nRT space space At space straight T space equals space 20 º straight C space equals space 293 space straight K space space For space benzene comma space straight P space equals space straight P subscript straight B superscript 0 space equals space nRT over straight V space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1 space cross times 0.082 space cross times 293 over denominator 244.58 end fraction space equals space 0.098 space atm space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 74.48 space torr space left parenthesis because space 1 space atm space equals space 760 space torr right parenthesis According space to space Raoult apostrophe straight s space law comma space straight P subscript straight B space equals space space straight P subscript straight B superscript 0 space straight X subscript straight B space equals space 74.48 space straight X subscript straight B space space straight P subscript straight T space equals straight P subscript straight T superscript 0 space space straight X subscript straight T space equals space 22.04 space left parenthesis 1 space – space straight X subscript straight B right parenthesis space space And space straight P subscript straight M space equals space straight P subscript straight B superscript 0 space straight X subscript straight B space plus space straight P subscript straight T superscript 0 space space straight X subscript straight T space or space 46.0 space equals space 74.48 space straight X subscript straight B space plus space 22.04 space left parenthesis 1 space – space straight X subscript straight B space right parenthesis space Solving comma space straight X subscript straight B space equals space 0.457 space According space to space Dalton apostrophe straight s space law comma space space straight P subscript straight B space equals space PM space space straight X subscript straight B superscript apostrophe space space left parenthesis In space vapour space phase right parenthesis space space or space mole space fraction space of space benzene space in space vapour space form comma straight X subscript straight B superscript apostrophe space equals space straight P subscript straight B over straight P subscript straight M space equals space fraction numerator 74.48 space cross times space 0.457 over denominator 46.0 end fraction space equals space bold 0 bold. bold 74   end style
Answered by Hanisha Vyas | 27 Jul, 2015, 12:23: PM
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