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The vapoure pressure of benzene and toluene at `20^(circ)C` are `75 mm` of `Hg` and `22mm` of `Hg` respectively. `23.4g` of benzene and `64.4g` of toluene are mixed. If two forms ideal solution, calculate the mole fraction of benzene in vapour phase when vapour are in equilibrium with liquid mixture. |
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Answer» Calculation of the total vapour pessure `"Moles of benzene "(n_(B))=("Mass of" CH_(3)COOC_(2)H_(5))/("Molar massof benzene")(C_(6)H_(6))=((23.4g))/((78g mol^(-1)))=0.3` mol `" Mole fraction of benzene"(x_(A))=("Mass of toluene")/("Molar mass jof toluene"(C_(7)H_(8)))= ((64.4g))/((92g mol^(-1)))=0.7` mol `"Mole faction of benzene"(x_(B))=n_(B)/(n_(a)+n_(B))=((0.3mol))/((0.3mol+0.7 mol))=0.3` `"mole fraction of toluene "(x_(A))=n_(A)/(n_(A)+n_(B))/((0.7 mol))/((0.3 mol+0.7mol))=0.7 ` `"Mole fraction of toluene"((x_(A))=n_(A)/((0.3mol+0.7mol))=0.7` `" Vopour pressre of benzene in soulution "(P_(B))=P_(B)^(@)xxx_(B)=(67mm)xx0.3=22.5mm` Calculoation of mnole fractin of benxene in the vapur phase. `"Mole fration of benzene in vapur phase"=("Partial lllvapour prassure of benzene")/("Total vapour pressure")` `((22.5mm))/((37.9mm))=0.59`. |
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