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

Question

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 (A) 0.635 MeOH , 0.365 EtOH (B) 0.173 MeOH , 0.827 EtOH (C) 0.574 MeOH , 0.326 EtOH (D) 0.365 MeOH , 0.635 EtOH

Tardigrade Admin · Accepted Answer

P Total =2.619+4.556=7.175 kPa Mole fraction in vapour phase =( text Partial pressure / text Total pressure ) X MeOH =(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 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

0.173 MeOH , 0.827 EtOH

0.574 MeOH , 0.326 EtOH

0.365 MeOH , 0.635 EtOH

$P_{T o t a l} = 2.619 + 4.556 = 7.175 k P a$

Mole fraction in vapour phase $= \frac{Partial pressure}{Total pressure}$

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

$\Rightarrow X_{EtO H} = 1 - 0.365 = 0.635$

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