One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is

Question

One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is (A) 2/3 (B) 1/3 (C) 1/2 (D) 3/2

Tardigrade Admin · Accepted Answer

(Po-PS/Po)=(1/1+2)=(1/3) ∴ 1-(PS/Po)=(1/3) Thus (Ps/Po)=1-(1/3)=(2/3)

Q. One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is

2/3

1/3

1/2

3/2

$\frac{P ^{o} - P _{S}}{P ^{o}} = \frac{1}{1 + 2} = \frac{1}{3}$
$∴ 1 - \frac{P _{S}}{P ^{o}} = \frac{1}{3}$
Thus $\frac{P _{s}}{P ^{o}} = 1 - \frac{1}{3} = \frac{2}{3}$

Q. One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is

2/3

1/3

1/2

3/2

PoPo−PS​​=1+21​=31​ ∴1−PoPS​​=31​ Thus PoPs​​=1−31​=32​

$\frac{P ^{o} - P _{S}}{P ^{o}} = \frac{1}{1 + 2} = \frac{1}{3}$
$∴ 1 - \frac{P _{S}}{P ^{o}} = \frac{1}{3}$
Thus $\frac{P _{s}}{P ^{o}} = 1 - \frac{1}{3} = \frac{2}{3}$