Question

An ideal solution of benzene and toluene boils at $1atm$ pressure at $90{}^{\circ }C$. At $90{}^{\circ }C$ benzene has a vapour pressure of 1022 mm of Hg and toluene has a vapour pressure of 406 mm of Hg. The mole fraction of benzene in the solution:

• A. $0.360$
• B. $0.640$
• C. $0.575$
• D. $0.425$

Answer 1

The correct option is C $0.575$

Given,
$\text{Vapour pressure of pure benzene},P_B^0=\text{1022 mm of Hg}$
$\text{Vapour pressure of pure toluene},P_T^0=\text{406 mm of Hg}$

We know,
$P_B^0{\chi }_B+P_T^0{\chi }_T=P_{total}$

$1022{\chi }_B+406(1-{\chi }_B)=1atm=\text{760 mm of Hg}$

$\therefore {\chi }_B=0.575$