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The vepour pressure of ethanol and mthanol are 44.5 mm and 88.7 mm of Hg respectively. A solution prepared by mixing 60 g of ethanol and 40 g of methanol. Assuming the solution to be idea, calculate the vepour pressure of the mixing 60 g of ethanol and 40 g of methanol. Assming to be ideal, calculate the vepour pressure of the solution. |
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Answer» `"No. of moles of methanol "(n_(B))=("Mass of" CH_(3)OH)/("Gram molar mass")=((40 g))/((32 gmol ^(-1)))=1.25 mol` `" No. of moles of ethanol "(n_(A))=("Mass of" C_(2)H_(5)OH)/("Gram molar mass")=((60g))/(46gmol^(-1))=1.30 mol ` `" Mole fraction of methanol"(x_(B))=n_(B)/(n_(B)+n_(A))=((1.30 mol))/((1.25 mol+1.30 mol))=0.51 ` `"Vapour pressure of pure methanol" (P_(B)^(@))=88.7 mm` `" Vapour pressure of pure ethanol "(P_(A)^(@))=44.5 mm` `"Partial vapour pressure of methanol" (P_(B)^(@))=P_(B)^(@)x_(B)=(88.7mmxx0.49)=43.46 mm` `"Partial vapour pressure of methanol "(P_(A)^(@))=P_(A)^(@)x_(A)=(44.5mmxx0.51)=22.70 mm` since the solution is ideal in nature, `" Total vepur pressure of solution" (P)=P_(A)+P_(B)=(22.70+43.46)=66.166mm` |
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