
Vapour pressure of pure water at 298 K is $23.8$ mmHg. 50 g of urea is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution at its relative lowering.
Answer
569.1k+ views
Hint: The vapour pressure of a solvent is the pressure exerted by the vapour of the solvent on the walls of the container and the addition of solute to the solvent reduces this vapour pressure. This relative lowering of the vapour pressure solution is dependent on the number of the solute molecules.
Formula used:
$\dfrac{{{{\text{P}}^{\text{0}}} - {\text{ P}}}}{{{{\text{P}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_1}}}$
Complete step by step answer:
The relative lowering of vapour pressure is a colligative property of the solution that is mathematically represented as:
$\dfrac{{{{\text{P}}^{\text{0}}} - {\text{ P}}}}{{{{\text{P}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_1}}}$
Here, ${{\text{w}}_2}$ = 50 g, Molecular weight of urea, ${{\text{M}}_2} = 60.06$ , ${{\text{w}}_1}$ = 850 g, ${{\text{M}}_1}$ = 18 ${\text{g/mol}}$ and the vapour pressure of the pure solvent ${{\text{P}}^0}$ = $23.8$ mmHg
Putting the values of the variable in the equation, we get,
$\dfrac{{23.8 - {\text{P}}}}{{23.8}}{\text{ = }}\dfrac{{50 \times 18}}{{60.6 \times 850}}$
$ \Rightarrow 23.8 - {\text{P = }}\dfrac{{{\text{21420}}}}{{51510}} = 0.415$
$ \Rightarrow {\text{ P = }}23.8 - 0.415$ = $23.385$ mmHg
Solving this we get, the final vapour pressure of the solution is $23.385$mm of Hg.
The relative lowering of vapour pressure is thus, $\dfrac{{{P^0} - P}}{{{P^0}}} = \dfrac{{\Delta P}}{{{P^0}}}$ = $\dfrac{{0.415}}{{23.8}}$ = \[0.0174\]
As this is a ratio of two quantities so the relative lowering has no unit.
Note:
The colligative properties of the solution depend on the number of the solute molecules present in the solution and not on the type or the nature of the solute.
Besides the lowering of vapour pressure, there are three other colligative properties namely, elevation in boiling point of the solvent, depression in freezing point of the solvent and osmotic pressure.
Formula used:
$\dfrac{{{{\text{P}}^{\text{0}}} - {\text{ P}}}}{{{{\text{P}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_1}}}$
Complete step by step answer:
The relative lowering of vapour pressure is a colligative property of the solution that is mathematically represented as:
$\dfrac{{{{\text{P}}^{\text{0}}} - {\text{ P}}}}{{{{\text{P}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_1}}}$
Here, ${{\text{w}}_2}$ = 50 g, Molecular weight of urea, ${{\text{M}}_2} = 60.06$ , ${{\text{w}}_1}$ = 850 g, ${{\text{M}}_1}$ = 18 ${\text{g/mol}}$ and the vapour pressure of the pure solvent ${{\text{P}}^0}$ = $23.8$ mmHg
Putting the values of the variable in the equation, we get,
$\dfrac{{23.8 - {\text{P}}}}{{23.8}}{\text{ = }}\dfrac{{50 \times 18}}{{60.6 \times 850}}$
$ \Rightarrow 23.8 - {\text{P = }}\dfrac{{{\text{21420}}}}{{51510}} = 0.415$
$ \Rightarrow {\text{ P = }}23.8 - 0.415$ = $23.385$ mmHg
Solving this we get, the final vapour pressure of the solution is $23.385$mm of Hg.
The relative lowering of vapour pressure is thus, $\dfrac{{{P^0} - P}}{{{P^0}}} = \dfrac{{\Delta P}}{{{P^0}}}$ = $\dfrac{{0.415}}{{23.8}}$ = \[0.0174\]
As this is a ratio of two quantities so the relative lowering has no unit.
Note:
The colligative properties of the solution depend on the number of the solute molecules present in the solution and not on the type or the nature of the solute.
Besides the lowering of vapour pressure, there are three other colligative properties namely, elevation in boiling point of the solvent, depression in freezing point of the solvent and osmotic pressure.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
State the principle of an ac generator and explain class 12 physics CBSE

Sketch the electric field lines in case of an electric class 12 physics CBSE

Derive an expression for electric potential at point class 12 physics CBSE

What is virtual and erect image ?

Explain the formation of energy bands in solids On class 12 physics CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

