Question

An ideal solution was obtained by mixing methanoland ethanol. If the partial vapour pressure of methanol and ethanol are 2.619 k Pa and 4.556 k the composition of vapour(in terms of mole fraction) will be

• 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 1

Answer: (B)

0.365 MeOH, 0.635 EtOH

Answer 2

Partial pressure of methanol $=2.619\, kP$ a Partial pressure of ethanol $=4.556 \,kPa$ Total pressure = Partial pressure of methanol $+$ Partial pressure of ethanol Total pressure $=2.619+4.556=7.175\, kP a$ $X_{\text {methanol: }}$ Mole fraction of methanol $X_{\text {ethanol: }}$ Mole fraction of ethanol The partial pressure of methanol $=X_{\text {methanol }} \times$ Total pressure $\therefore X _{\text {methanol }}=\frac{2.619}{7.175}$ $=0.365$ $\therefore$ Mole fraction of ethanol= $1-$ Mole fraction of methanol $=1-0.365$ $=0.635$