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The molar volume of liquid benzene (density = "0.877 g mL"^(-1)) increases by a factor of 2750 as it vaporises at 20^(@)C and that of liquid toluene (density "0.867 g mL"^(-1)) 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 the mole fraction of benzene in the vapour above the solution. |
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Answer» Solution :In vapour phase, `"1 MOLE of benzene, i.e., 78 g has volume at "20^(@)C=(78)/(0.877)xx"2750 mL"` `"1 mole of toluene, i.e., 92 g has volume at "20^(@)C=(92)/(0.867)xx"7720 mL"` `"Applying , PV = N RT"` `"For benzene vapour, "(P^(@))/(760)"(atm)"xx(78xx2750)/(0.877xx1000)(L)=1xx"0.0821 L atm K"^(-1)"mol"^(-1)xx293 K` `"or"P_(B)^(@)=74.74mm` `"For toluene vapour, "(P^(@))/(760)xx(92xx7720)/(0.867xx1000)=1xx0.0821xx293` `"or"P_(T)^(@)=22.37mm` `""P_("total")=x_(B)xxP_(B)^(@)+x_(T)xxP_(T)^(@)=x_(B)P_(B)^(@)+(1-x_(B))xxP_(T)^(@)` `THEREFORE""46=x_(B)=(74.74)+(1-x_(B))(22.37)` This on solving gives `x_(B)=0.45` (in the liquid phase) `therefore""x_(T)=1-0.45=0.55` (in the liquid phase) `"In vapour phase,"P_(B)=x_(B)P_(B)^(@)=0.45xx74.74=36.63mm` `""P_(T)=x_(T)P_(T)^(@)=0.55xx22.37=12.30 mm` `"Mole fraction of benzene in vapour phase "(y_(B))=(P_(B))/(P_(B)+P_(T))(36.63)/(36.63+12.30)=0.75` |
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