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Vapour pressures of benzene and toluene in a mixute at 50^(@)C are given in mm by P=179 X_(B)+92 where X_(B) is the mole fraction of benzene. Calculate. (b) Vapour pressure of a liquid mixture obtained by mixing 224 g benzene and 184 g of toluene. |
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Answer» Solution :(b) 224 g benzene `(C_(6)H_(6))=(224)/(78)" MOL" = "3 mol,184 g toluene "(C_(6)H_(5)CH_(3))=(184)/(92)=" 2 mol"` `X_(B)=(3)/(5)=0.6,""X_(T)=(2)/(5)=0.4` `P_(B)=0.6xx"271 mm"="162.6 mm," P_(T)=0.4xx92 = "36.8 mm", P_("Total")=162.6+36.8="199.4 mm"` |
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