Question

The vapor pressure of water at room temperature is $23.8$ mm of $Hg$ . The vapor pressure of an aqueous solution of sucrose with mole fraction $0.1$ is equal to:
A. $23.9$ mm of $Hg$
B. $24.2$ mm of $Hg$
C. $21.42$ mm of $Hg$
D. $31.44$ mm of $Hg$

Answer 1

Hint: The vapor pressure decreases as the temperature decreases. The vapor pressure of a solution is lower than that of the solvent. Thus, the vapor pressure of a solution will be equal to that of the solid at a lower temperature than in the case of the pure solvent.

Complete step by step answer:
As per the question,
The vapor pressure of water = ${P_o} = 23.8mmHg$
The vapor pressure of solution = ${P_s} = ?$
According to the lowering of vapor pressure, the vapor pressure of a solvent decreases when a solute is added in it. Mathematically, the lowering of vapor pressure is written as:
$\Delta P = {P_o} - {P_s}$
The lowering in vapor pressure with respect to the vapor pressure of the pure solvent is known as the relative lowering of vapor pressure. It is directly proportional to the mole fraction of the solute (in this case, sucrose). As it is a colligative property, it can be written mathematically as:
$\dfrac{{{P_o} - {P_s}}}{{{P_o}}} = i{x_B}$ ….. (i)
Where, ${P_o} = $ vapor pressure of the water = $23.8mmHg$
${P_s} = $ vapor pressure of the water and sucrose solution = $?$
$i = $ Van’t Hoff factor
For sucrose, $i = 1$
${x_B} = $ mole fraction of sucrose (solute)
Substituting the values in equation (i), we have:
$ \Rightarrow \dfrac{{23.8 - {P_s}}}{{23.8}} = 1 \times 0.1$
${P_s} = 21.42mmHg$
Thus, the correct option is C. $21.42$ mm of $Hg$ .


Note:
A reduction in the saturated vapor pressure of a pure liquid when a solute is introduced is known as the lowering of vapor pressure. If the solute is a solid of low vapor pressure, the decrease in the vapor pressure of the liquid is proportional to the concentration of particles of solute; i.e. to the number of dissolved molecules or ions per unit volume.