Question

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