Question

At 25${}^0{\text{C}}$, the vapour pressure of pure water is 25.0 mm Hg and that of an aqueous dilute solution of urea is 20 mm Hg. Calculate the molality of that solution.

Answer 1

Hint- Here, we will proceed by using Raoult’s Law to find the mole fractions of pure water and urea solution. Then, we will be using the relationship between molality and mole fractions to get the molality of the given urea solution.

Complete answer:
Formulas Used- Vapour pressure of the solution = (Vapour pressure of the pure solvent)$\times$(Mole fraction of the pure solvent), ${{\text{X}}_{{\text{pure solvent}}}} + {{\text{X}}_{{\text{solution}}}} = 1$ and Molality of the solution = $\dfrac{{\left( {{\text{Mole fraction of solution}}} \right) \times 1000}}{{\left( {{\text{Mole fraction of pure solvent}}} \right) \times \left( {{\text{Molar mass of pure solvent}}} \right)}}$.
Given, Vapour pressure of pure water, ${{\text{P}}_{{\text{water}}}}$= 25 mm Hg
Vapour pressure of an aqueous dilute solution of urea, ${{\text{P}}_{{\text{urea solution}}}}$= 20 mm Hg
Temperature = 25${}^0{\text{C}}$
According to Raoult’s Law, we can write
Vapour pressure of the solution = (Vapour pressure of the pure solvent)$\times$(Mole fraction of the pure solvent) $\to (1)$
Let the mole fractions of water and that of aqueous dilute solution of urea be ${{\text{X}}_{{\text{water}}}}$ and ${{\text{X}}_{{\text{urea solution}}}}$ respectively.
In this case, pure solvent is water and the solution is an aqueous dilute solution of urea
Using the formula given by equation (1), we can write
${{\text{P}}_{{\text{urea solution}}}} = {{\text{P}}_{{\text{water}}}} \times {{\text{X}}_{{\text{water}}}}$
By substituting ${{\text{P}}_{{\text{water}}}}$= 25 mm Hg and ${{\text{P}}_{{\text{urea solution}}}}$= 20 mm Hg in the above equation, we have

$$\Rightarrow {\text{20}} = {\text{25}} \times {{\text{X}}_{{\text{water}}}} \\\ \Rightarrow {{\text{X}}_{{\text{water}}}} = \dfrac{{{\text{20}}}}{{25}} = \dfrac{{\text{4}}}{5} \\\ \Rightarrow {{\text{X}}_{{\text{water}}}} = 0.8{\text{ }} \to {\text{(2)}} \\\$$

So, the mole fraction of water (pure solvent) is 0.8
Also, we know that the sum of the mole fractions of the pure solvent and that of the solution is always equal to 1.
i.e., ${{\text{X}}_{{\text{pure solvent}}}} + {{\text{X}}_{{\text{solution}}}} = 1$
${{\text{X}}_{{\text{water}}}} + {{\text{X}}_{{\text{urea solution}}}} = 1 \\\ \Rightarrow {{\text{X}}_{{\text{water}}}} = 1 - {{\text{X}}_{{\text{urea solution}}}}{\text{ }} \to {\text{(3)}} \\\$
By substituting equation (3) in equation (2), we get

$$\Rightarrow 1 - {{\text{X}}_{{\text{urea solution}}}} = 0.8 \\\ \Rightarrow {{\text{X}}_{{\text{urea solution}}}} = 1 - 0.8 \\\ \Rightarrow {{\text{X}}_{{\text{urea solution}}}} = 0.2 \\\$$

Since, the molar mass of water (${{\text{H}}_2}{\text{O}}$), M = 2(Molar mass of hydrogen) + Molar mass of oxygen = 2(1) + 16 = 18
According to the relation between molality and mole fraction, we can write
Molality of the solution = $\dfrac{{\left( {{\text{Mole fraction of solution}}} \right) \times 1000}}{{\left( {{\text{Mole fraction of pure solvent}}} \right) \times \left( {{\text{Molar mass of pure solvent}}} \right)}}$
Using the above formula, we can write
Molality of the urea solution = $\dfrac{{\left( {{{\text{X}}_{{\text{urea solution}}}}} \right) \times 1000}}{{\left( {{{\text{X}}_{{\text{water}}}}} \right) \times \left( {\text{M}} \right)}} = \dfrac{{\left( {{\text{0}}{\text{.2}}} \right) \times 1000}}{{\left( {{\text{0}}{\text{.8}}} \right) \times \left( {{\text{18}}} \right)}} = \dfrac{{2000}}{{144}} = 13.89$ m

Therefore, the molality of the given an aqueous dilute solution of urea is 13.89 m.
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Note- Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is m or mol/kg. Raoult’s Law states that a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.