A mixture of methanol (MeOH) and ethanol (EtO | vedclass

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(B) In the vapor phase,the mole fraction of a component is given by the ratio of its partial pressure to the total pressure of the solution.Total pres

Vedclass · Accepted Answer

$0.365 \ MeOH, 0.635 \ EtOH$

Vedclass · Answer

(B) In the vapor phase,the mole fraction of a component is given by the ratio of its partial pressure to the total pressure of the solution.Total pressure $P_{total} = P_{MeOH} + P_{EtOH} = 2.619 \ kPa + 4.556 \ kPa = 7.175 \ kPa$.Mole fraction of $MeOH$ in vapor phase $(y_{MeOH}) = \frac{P_{MeOH}}{P_{total}} = \frac{2.619}{7.175} = 0.365$.Mole fraction of $EtOH$ in vapor phase $(y_{EtOH}) = \frac{P_{EtOH}}{P_{total}} = \frac{4.556}{7.175} = 0.635$.

$A$ mixture of methanol $(MeOH)$ and ethanol $(EtOH)$ forms an ideal solution. If the partial pressures of methanol and ethanol are $2.619 \ kPa$ and $4.556 \ kPa$ respectively,what will be the mole fractions of the components in the vapor phase?

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