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 vapour (in terms of mole fraction) will be

Options

A. `0.635 MeOH`,`0.365 EtOH`
B. `0.365 MeOH`,`0.635 EtOH`
C. `0.574 MeOH`,`0.326 EtOH`
D. `0.173 MeOH`,`0.827 EtOH`

Answer

B

Explanation

`P_(EtOH)=P_(EtOH)^(@) xx "Mole fraction of EtOH in liquid phase"`
=`P_(T) xx "Mole fraction of EtOH in vapour phase"`
Thus, mole fraction of `EtOH` in vapour phase
=`P'/P_(T)=4.556/(4.556 + 2.619) = 0.635`