Question

The vapour pressure of a dilute aqueous solution of glucose is ${\text{750 mm Hg}}$ at ${\text{373 K}}$. Calculate the molality and mole fraction of the solute.

Answer 1

Molality of a solution is the ratio of the moles of solute to the weight of the solvent (formula given). Mole fraction of the solute is the ratio of moles of a solute to the total number of moles of solute and solvent in the solution (formula given). We shall find out the mole fraction of the solute using the colligative property of relative lowering of vapour pressure (formula given). From that, we shall derive a relation between mole fraction and molality to find the molality of the solution.
Formula used:
${{\text{X}}_{\text{A}}}{\text{ = }}\dfrac{{{{\text{P}}^{\text{o}}}{\text{ - }}{{\text{P}}_{\text{S}}}}}{{{{\text{P}}^{\text{o}}}}}$ where ${{\text{X}}_{\text{A}}}$ is the mole fraction of the solute, ${{\text{P}}^{\text{o}}}$ is the vapour pressure of pure solvent, ${{\text{P}}_{\text{S}}}$ is the vapour pressure of the solution. (Eq. 1)
${\text{Molality = }}\dfrac{{{{\text{n}}_A}{\text{} \times 1000}}}{{{{\text{W}}_B}{\text{ (in g)}}}}$ where ${{\text{n}}_{\text{A}}}$ is the number of moles of solute in the solution, ${{\text{W}}_{\text{B}}}$ is the weight of the solvent. (Eq. 2)
${{\text{X}}_{\text{A}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{A}}}{\text{ + }}{{\text{n}}_{\text{B}}}}}$ where ${{\text{X}}_{\text{A}}}$ is the mole fraction of the solute, ${{\text{n}}_{\text{A}}}$ is the number of moles of solute in the solution, ${{\text{n}}_{\text{B}}}$ is the number of moles of the solvent, and (Eq. 3)

Complete step by step answer:
As we already know that, the boiling point of water is ${\text{10}}{{\text{0}}^o}{\text{C = 373 K}}$ , so the vapour pressure of water at ${\text{373 K}}$ will be ${\text{1 atm = 760 mm Hg}}$ . Now, we know both the vapour pressure of pure solvent, i.e. water and also the vapour pressure of the solution. So, substituting these values in Eq. 1, we get:
${{\text{X}}_{\text{A}}}{\text{ = }}\dfrac{{{\text{760 - 750}}}}{{760}}$
$\Rightarrow {{\text{X}}_{\text{A}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{76}}}}{\text{ = 0}}{\text{.013}}$
As it is given in the question that the solution used is a dilute solution. So, Eq. 3 can also be written as:
${{\text{X}}_{\text{A}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{B}}}}}$ , this is because${{\text{n}}_{\text{A}}} \ll {{\text{n}}_{\text{B}}}$ . (Eq. 4)
Now, according to mole concept:
${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
$\Rightarrow {\text{mass = moles} \times \text{molar mass}}$
Using this above relation in Eq. 2 for ${{\text{W}}_{\text{B}}}$ we get:
${\text{Molality = }}\dfrac{{{{\text{n}}_{\text{A}}}{\text{} \times 1000}}}{{{{\text{n}}_{\text{B}}}{ \times \text{molar mass of water (in g)}}}}$
Comparing this with Eq. 4, we can conclude that:
${\text{Molality = }}{{\text{X}}_{\text{A}}}{\text{} \times }\dfrac{{{\text{1000}}}}{{{\text{molar mass of water}}}}$ this is because, ${{\text{X}}_{\text{A}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{B}}}}}$.
We already know that the molar mass of water is ${\text{18 g mo}}{{\text{l}}^{{\text{ - 1}}}}$ . So, after substituting the appropriate values in the above equation, we get:
${\text{Molality = 0}}{\text{.013} \times }\dfrac{{{\text{1000}}}}{{18}}$
$\Rightarrow {\text{Molality = 0}}{\text{.722 m}}$
$\therefore$ The molality of the given dilute aqueous solution of glucose is ${\text{0}}{\text{.722 m}}$ and the mole fraction of solute in the solution is ${\text{0}}{\text{.013}}$ .

Note:
The vapour pressure of any volatile liquid at its boiling point will always be equal to ${\text{1 atm = 760 mm Hg}}$. In this question, we could derive the relation between mole fraction and molality only because the solution was diluted. If the solution was concentrated, we would have to apply other methods to find the molality of the solute.