Question

An ideal solution of hexane and heptane at $30{ }^{\circ} C$ has a vapour pressure of $95$ bar with hexane mole fraction $0.305 .$ In vapour phase hexane mole fraction is $0.555 .$ The vapour pressure of pure hexane and heptane at $30^{\circ} C$ respectively in bar are

• A. 172.9,60.9
• B. 60.8,172.9
• C. 30.4,86.5
• D. 86.5,30.4

Answer 1

Answer: (A)

Given, vapour pressure of ideal solution of hexane $\left(p_{T}\right)=95$ bar

Mole fraction of hexane $\left(\chi_{A}\right)=0.305$

So, $\left(\chi_{B}\right)=1-0.305=0.695$

Mole fraction of vapour phase hexane $\left(Y_{A}\right)=0.555$

and of heptane $\left(Y_{B}\right)=1-0.555=0.445$

From Raoult's law,

Vapour pressure of hexane $=p_{A}^{\circ}$

Vapour pressure of heptane $=p_{B}^{\circ}$

$p_{T}\left(Y_{A}\right)=p_{A}^{\circ}\left(\chi_{A}\right), p_{T}\left(Y_{B}\right)=p_{B}^{\circ}\left(\chi_{B}\right)$

$95(0.555)=p_{A}^{\circ}(0.305), 95(0.445)=p_{B}^{\circ}(0.695)$

$p_{A}^{\circ}=172.9$ bar, $p_{B}^{\circ}=60.9$ bar