The vapour pressure of ethanol (C_2H_5OH) and | vedclass

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(A) $1$. Calculate the moles of each component: $n_{	ext{ethanol}} = \frac{60 \ g}{46 \ g/mol} \approx 1.304 \ mol$ $n_{	ext{methanol}} = \frac{40 \

Vedclass · Accepted Answer

$0.656$

Vedclass · Answer

(A) $1$. Calculate the moles of each component: $n_{	ext{ethanol}} = \frac{60 \ g}{46 \ g/mol} \approx 1.304 \ mol$ $n_{	ext{methanol}} = \frac{40 \ g}{32 \ g/mol} = 1.25 \ mol$ $2$. Calculate the mole fractions in the liquid phase: $x_{	ext{ethanol}} = \frac{1.304}{1.304 + 1.25} \approx 0.510$ $x_{	ext{methanol}} = \frac{1.25}{1.304 + 1.25} \approx 0.490$ $3$. Calculate the partial pressures using Raoult's Law: $P_{	ext{ethanol}} = x_{	ext{ethanol}} 	imes P^0_{	ext{ethanol}} = 0.510 	imes 44.5 \ mm \ Hg \approx 22.695 \ mm \ Hg$ $P_{	ext{methanol}} = x_{	ext{methanol}} 	imes P^0_{	ext{methanol}} = 0.490 	imes 88.7 \ mm \ Hg \approx 43.463 \ mm \ Hg$ $4$. Calculate the total pressure: $P_{	ext{total}} = 22.695 + 43.463 = 66.158 \ mm \ Hg$ $5$. Calculate the mole fraction of methanol in the vapour state $(y_{	ext{methanol}})$: $y_{	ext{methanol}} = \frac{P_{	ext{methanol}}}{P_{	ext{total}}} = \frac{43.463}{66.158} \approx 0.657$ The closest value is $0.656$.

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