The vapour pressure of a solvent decreases by 10 mm of Hg than a non-volatile solute was added is 0.2 . What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of Hg.

Question

The vapour pressure of a solvent decreases by 10 mm of Hg than a non-volatile solute was added is 0.2 . What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of Hg. (A) 0.6 (B) 0.4 (C) 0.2 (D) 1

Tardigrade Admin · Accepted Answer

P °- P =10 Apply, ( P °- P s / P °)= x B (10/ P °)=0.2 ⇒ P °=50 Then, P °- P s =20 (P0-Ps/P0)=xB ⇒ (20/50)=xB=0.4 (∵ xA+xB=1) ∴ xA=0.6

Q. The vapour pressure of a solvent decreases by $10 mm$ of $H g$ than a non-volatile solute was added is $0.2.$ What should be the mole fraction of the solvent if the decrease in vapour pressure is to be $20 mm$ of $H g$ .

0.6

0.4

0.2

1

$P^{\circ} - P = 10$

Apply, $\frac{P ^{\circ} - P _{s}}{P ^{\circ}} = x_{B}$

$\frac{10}{P ^{\circ}} = 0.2 \Rightarrow P^{\circ} = 50$

Then, $P^{\circ} - P_{s} = 20$

$\frac{P ^{0} - P _{s}}{P ^{0}} = x_{B}$

$\Rightarrow \frac{20}{50} = x_{B} = 0.4$

$(∵ x_{A} + x_{B} = 1)$

$∴ x_{A} = 0.6$

Q. The vapour pressure of a solvent decreases by 10mm of Hg than a non-volatile solute was added is 0.2. What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20mm of Hg.

0.6

0.4

0.2

1

P∘−P=10 Apply, P∘P∘−Ps​​=xB​ P∘10​=0.2⇒P∘=50 Then, P∘−Ps​=20 P0P0−Ps​​=xB​ ⇒5020​=xB​=0.4 (∵xA​+xB​=1) ∴xA​=0.6

Q. The vapour pressure of a solvent decreases by $10 mm$ of $H g$ than a non-volatile solute was added is $0.2.$ What should be the mole fraction of the solvent if the decrease in vapour pressure is to be $20 mm$ of $H g$ .

$P^{\circ} - P = 10$

Apply, $\frac{P ^{\circ} - P _{s}}{P ^{\circ}} = x_{B}$

$\frac{10}{P ^{\circ}} = 0.2 \Rightarrow P^{\circ} = 50$

Then, $P^{\circ} - P_{s} = 20$

$\frac{P ^{0} - P _{s}}{P ^{0}} = x_{B}$

$\Rightarrow \frac{20}{50} = x_{B} = 0.4$

$(∵ x_{A} + x_{B} = 1)$

$∴ x_{A} = 0.6$