Question

X is a non-volatile solute and Y is volatile solvent. The following vapor pressures are obtained by dissolving X in Y.

$X/mol{(L)^{ - 1}}$	Y/mm Hg
$0.1$	${P_1}$
0.25	${P_2}$
0.01	${P_3}$

The correct order of vapour pressure is:
A.${P_1} < {P_2} < {P_3}$
B.${P_3} < {P_2} < {P_1}$
C.${P_3} < {P_1} < {P_2}$
D.${P_2} < {P_1} < {P_3}$

Answer 1

The vapor pressure can be defined as the pressure exerted by vapor. The volatile solvent has a high vapor pressure. The addition of a solute to the volatile solvent results in the decrease of vapor pressure of a volatile solvent as the solute particles added to the solvent increase in the boiling point results in the decreasing of vapor pressure.

Complete answer:
Given that X is a non-volatile solute which means the boiling point is not much high. The Y is a non-volatile solvent with high vapor pressure and has a high boiling point. By dissolving X in Y the boiling point of Y will be increased, resulting in the decreasing of vapor pressure. Thus, higher concentration of solute means a higher amount of solute is added to the volatile solvent. Thus, the higher concentration results in the lowering of vapor pressure.
In the given concentrations of X $0.25$ the highest concentration has lower vapor pressure. $0.1$is the intermediate concentration that has the intermediate vapor pressure. $0.01$ is the lower concentration and has the highest vapor pressure.

Thus, ${P_2} < {P_1} < {P_3}$ is the correct order and option D is the correct one.

Note:
The solute addition to the solvent results in the high concentration of solution, the concentration is the amount of solute dissolved in the solvent. Vapor pressure depends upon the concentration. Vapor pressure is one of the colligative properties that does not depend upon the nature of the solute.