Methanol and ethanol from nearly an ideal solution. If the vapour pressure of pure methanol and ethanol at `350 K are 8.1xx10^(4)and 4.5xx10^(4)Nm^(-2)` respeclate the total vapour pressure of the solution and the mole fraction of methanol in the vapour phase if the solution contains 64 g of methanol and 46 g of ethanol at this temperature.

Explanation

Moles of etanol `(n_(B)`=`(Mass of ethanol)/(Molar mass of ethanol(C_(2)H_(5)OH))=((46g))/((46gmol^(-1)))=1 mol`
Moles of methanol `(n_(2))`=`(Mass of methanol)/(Molar mass of methanol(CH_(3)OH))=1/3`
Mole fraction of methanol `(x_(B))`=`n_(A)/(b_(A)+n_(B))=((2 mol))/((1 moil+ 2mol))=2/3`
Mole fractin of methanol (x_(A))= `n_(A)/(n_(A)+n_(B))=((2 mol))/((1 mol + 2mol))=2/3`
`"Vapour pressure of ethanol in mixtue "(P_(B))=P_(B)^(@)xx x_(B)=(4.5xx10^(4)Nm^(-2))xx1/3`
=`1.5xx10^(4) Nm^(-2)`
`"Vapour pressure of methanol in mixture "((P_(A))=P_(A)^(@)xx x_(A))=(8.1xx10^(4) Nm^(-2))xx 2/3`
=`5.4xx10^(4) Nm^(-2)`
`"Total vepour pressure of mixture"=(1.5xx10^(4)+5.4xx10^(4))=6.9xx10^(4)Nm^(-2)`
`"Mole fraction of methanol in the vepour phase" (x_(CH_(3)OH))=P_(A)/(P_(A)+P_(B))=((5.4xx10^(4)Nm^(-2)))/((6.9xx10^(4)Nm^(-2)))`
=0.7826