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(B) Partial pressure of methanol $= 2.619 \, kPa$Partial pressure of ethanol $= 4.556 \, kPa$Total pressure $= 	ext{Partial pressure of methanol} + \

Vedclass · Accepted Answer

$0.365$ methanol,$0.635$ ethanol

Vedclass · Answer

(B) Partial pressure of methanol $= 2.619 \, kPa$Partial pressure of ethanol $= 4.556 \, kPa$Total pressure $= 	ext{Partial pressure of methanol} + 	ext{Partial pressure of ethanol}$Total pressure $= 2.619 + 4.556 = 7.175 \, kPa$Let $y_{methanol}$ be the mole fraction of methanol in the vapour phase.$y_{methanol} = \frac{P_{methanol}}{P_{total}} = \frac{2.619}{7.175} = 0.365$Let $y_{ethanol}$ be the mole fraction of ethanol in the vapour phase.$y_{ethanol} = \frac{P_{ethanol}}{P_{total}} = \frac{4.556}{7.175} = 0.635$Thus,the composition of the vapour is $0.365$ methanol and $0.635$ ethanol.Hence,the correct option is $B$.

An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are $2.619 \, kPa$ and $4.556 \, kPa$ respectively,the composition of the vapour (in terms of mole fraction) will be

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