Question

Vapour pressure in mm Hg of 0.1 mole of urea in 180 g of water at ${25^o}{\text{C}}$ is (The Vapour pressure of water at ${25^0}{\text{C}}$ is 24 mm Hg)
A.\[\;2.376\]
B.\[20.76\]
C.\[23.76\]
D.\[24.76\]

Answer 1

Hint:The question can be solved from the knowledge of relative lowering of vapour pressure due to the addition of solute particles to the solvent. This relative lowering of vapour pressure is a colligative property of the solution.
Formula Used: $\dfrac{{{{\text{P}}^{\text{0}}} - {\text{ P}}}}{{{{\text{P}}^{\text{0}}}}}{\text{ = }}\dfrac{{{{\text{w}}_{\text{2}}} \times {{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}} \times {{\text{w}}_1}}}$
where, ${{\text{P}}^0}$ is the vapour pressure of the pure solvent, ${\text{P}}$ is the vapour pressure of the solution, ${{\text{w}}_2}$ is the weight of the solute in g, ${{\text{M}}_2}$ is the molar mass of the solute, ${{\text{w}}_1}$ is the weight of water and ${{\text{M}}_1}$ is the molecular mass of the solvent.

Complete step by step answer:
The relative lowering of vapour pressure can be 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, $\dfrac{{{{\text{w}}_2}}}{{{{\text{M}}_2}}}$ = number of moles = $0.1$ mole of Urea,
${{\text{w}}_1}$ = 180 g, ${{\text{M}}_1}$ = 18 ${\text{g/mol}}$ and the vapour pressure of the pure solvent ${{\text{P}}^0}$= $24$ mmHg
Putting the values of the variable sin the equation, we get,
$\dfrac{{24 - {\text{P}}}}{{24}}{\text{ = }}\dfrac{{0.1 \times 18}}{{{\text{ }}180}}$
$ \Rightarrow 24 - {\text{P = }}\dfrac{{{\text{0}}{\text{.1}} \times {\text{24}} \times {\text{18}}}}{{180}} = \dfrac{{43.2}}{{180}}$
$ \Rightarrow {\text{P}} = {\text{ }}24 - 0.24 = 23.76$

So, the correct answer is option C, $23.76$ mm of Hg.

Note:
The theory behind the lowering of the vapour pressure of the solvent due to the addition of the solute is that, when the solvent is pure then the vapour phase contains only solvent molecules and after the addition of the solute particles the solvent molecules are bound to the solute molecules by certain attractive forces and hence at the same temperature, lesser number of molecules are available and the vapour pressure of the solution falls down.