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

Question

An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol in solution are 2.619 kPa and 4.556 kPa respectively, 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

Tardigrade Admin · Accepted Answer

P text Total =2.619+4.556=7.175 kPa X MeOH =(P MeOH /P text Total )=(2.619/7.175)=0.365 X text EtOH =1-0.365=0.635

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

0.635 MeOH , 0.365 EtOH

0.365 MeOH , 0.635 EtOH

0.574 MeOH , 0.326 EtOH

0.173 MeOH , 0.827 EtOH

$P_{Total} = 2.619 + 4.556 = 7.175 k P a$

$X_{M e O H} = \frac{P _{M e O H}}{P _{Total}} = \frac{2.619}{7.175} = 0.365$

$X_{EtOH} = 1 - 0.365 = 0.635$

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