At 40° C, the vapour pressure of pure liquids, benzene and toluene, are 160 mmHg and 60 mmHg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be

Question

At 40° C, the vapour pressure of pure liquids, benzene and toluene, are 160 mmHg and 60 mmHg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be (A) 140 mmHg (B) 110 mmHg (C) 220 mmHg (D) 100 mmHg

Tardigrade Admin · Accepted Answer

For ideal solution, p text total =p1° x1+p2° x2 =160 × (1/2)+60 × (1/2)=110 mmHg

Q. At $4 0^{\circ} C$ , the vapour pressure of pure liquids, benzene and toluene, are $160 mm H g$ and $60 mm H g$ respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be

$140 mm H g$

$110 mm H g$

$220 mm H g$

$100 mm H g$

For ideal solution, $p_{total} = p_{1}^{\circ} x_{1} + p_{2}^{\circ} x_{2}$
$= 160 \times \frac{1}{2} + 60 \times \frac{1}{2} = 110 mm H g$

Q. At 40∘C, the vapour pressure of pure liquids, benzene and toluene, are 160mmHg and 60mmHg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be

140mmHg

110mmHg

220mmHg

100mmHg