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The vapour pressure of ethanol and metha...

The vapour pressure of ethanol and methanol are `44.0 mm` and `88.0 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60 g` of ethanol with `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Text Solution

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Mol. Mass of ethyl alcolhol `(C_(2)H_(5)OH) = 46`
Number of moles of ethylalcohol `= (60)/(46) = 1.304`
Mol. Mass of methyl alcohol `(CH_(3)OH)=32`
Number of moles of methyl alcohol `= (40)/(32) = 1.25`
Mole fraction of ethyl alcohol , `chi_(A)=(1.304)/(1.304+1.25)`
`=0.5107 mm Hg`
Mole fraction of methyl alcohol ,`chi_(B)=(1.25)/(1.304+1.25)`
`=0.4893 mm Hg`
Partial pressure of ethyl alcohol `= chi_(A)P_(A)^(@)`
`=22.47 mm Hg`
Partial pressure of ethyl alcohol `chi_(B)P_(B)^(@)`
`= 43.05 mm Hg`
Total vapour pressure of solution `= 22.47 + 43.05`
`65.52 mm Hg`
Mole fraction of methyl alcohol in the vapour
`=("Partial pressure of" CH_(3)OH)/("Total vapour pressure" ) =(43.05)/(65.52) =0.6570 `
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