X is a non-volatile solute and Y is a volatile solvent. The following vapour pressures are observed by dissolving X in Y. X / mol L -1 Y / mm of Hg 0.10 P 1 0.25 P 2 0.01 P3 The correct order of vapour pressures is

Question

X is a non-volatile solute and Y is a volatile solvent. The following vapour pressures are observed by dissolving X in Y. X / mol L -1 Y / mm of Hg 0.10 P 1 0.25 P 2 0.01 P3 The correct order of vapour pressures is (A) p1 < p2 < p3 (B) p3 < p2 < p1 (C) p3 < p1 < p2 (D) p2 < p1 < p3

Tardigrade Admin · Accepted Answer

When a non-volatile solute is added to a volatile solvent, the solute covers up some of the surface of solvent. Thus, less surface area is available for vaporisation of solvent and hence, vapour pressure decreases. As the amount of non-volatile solute increases, vapour pressure decreases accordingly.

∵

The order of concentration of

X

is

0.01 < 0.10 < 0.25

∴

The order of vapour pressure is

p_{3} > p_{1} > p_{2}

$X / m o l L^{- 1}$	$Y / mm$ of $H g$
0.10	$P_{1}$
0.25	$P_{2}$
0.01	$P_{3}$

Q. $X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ .

$X / m o l L^{- 1}$ $Y / mm$ of $H g$

0.10 $P_{1}$

0.25 $P_{2}$

0.01 $P_{3}$

The correct order of vapour pressures is

$p_{1} < p_{2} < p_{3}$

$p_{3} < p_{2} < p_{1}$

$p_{3} < p_{1} < p_{2}$

$p_{2} < p_{1} < p_{3}$

Q. X is a non-volatile solute and Y is a volatile solvent. The following vapour pressures are observed by dissolving X in Y. X/molL−1 Y/mm of Hg 0.10 P1​ 0.25 P2​ 0.01 P3​ The correct order of vapour pressures is

p1​<p2​<p3​

p3​<p2​<p1​

p3​<p1​<p2​

p2​<p1​<p3​

Q. $X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ .

$X / m o l L^{- 1}$ $Y / mm$ of $H g$

0.10 $P_{1}$

0.25 $P_{2}$

0.01 $P_{3}$

The correct order of vapour pressures is

$p_{1} < p_{2} < p_{3}$

$p_{3} < p_{2} < p_{1}$

$p_{3} < p_{1} < p_{2}$

$p_{2} < p_{1} < p_{3}$