Question

\[6g\] of urea (molecular weight $ = 60$) was dissolved in $9.9$ moles of water. If the vapour pressure of pure water is ${P^ \circ }$, the vapour pressure of the solution is:
A.$0.10{P^ \circ }$
B.$1.10{P^ \circ }$
C.$0.90{P^ \circ }$
D.$0.99{P^ \circ }$

Answer 1

Hint: Mole fraction of solute is defined as the ratio of moles of solute to the total number of moles present in the solution while mole fraction of solvent is defined as the ratio of moles of solvent to the total number of moles present in the solution.

Complete step by step solution:
First of all we will discuss the colligative properties.
Colligative properties: Those properties of solution which depend only on the ratio of solute particles to the number of solvent particles, are known as colligative properties. There are four colligative properties:
1.Osmotic pressure
2.Decrease in freezing point
3.Elevation in boiling point
4.Relative lowering in vapour pressure
Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute.
Osmotic pressure: It is defined as the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.
Relative lowering in vapour pressure: When non- volatile solutes are added to a solution then there is a decrease in its vapour pressure, which is known as relative lowering in vapour pressure.
The question is based on relative lowering in vapour pressure. The relative lowering in vapour pressure of the solution is equal to the mole fraction of solute. Here we are given with \[6g\] of urea (molecular weight $ = 60$) was dissolved in $9.9$ moles of water and the vapour pressure of pure water is ${P^ \circ }$. And the relation between vapour pressure of pure solvent with that of solution is as:
$P = {P^ \circ }\dfrac{{{n_2}}}{{{n_1} + {n_2}}}$ where $P$ is vapour pressure of solution, ${P^ \circ }$ is vapour pressure of pure solvent, ${n_2}$ number of moles of solvent and ${n_1}$ is the number of moles of solute.
In the question, ${n_2} = 9.9,{n_1} = \dfrac{6}{{60}} = 0.1$
Putting these values in the equation we get,
$
\Rightarrow P = {P^ \circ }\dfrac{{9.9}}{{9.9 + 0.1}} \\
\Rightarrow P = {P^ \circ }\dfrac{{9.9}}{{10}} \\
\Rightarrow P = {P^ \circ } \times 0.99 \\
\Rightarrow P = 0.99{P^ \circ } \\
 $

Hence option D is correct.

Note: The formula of Relative lowering in vapour pressure comes from Raoult’s law which states that vapour pressure of any component in a solution is equal to the mole fraction of that component multiplied by the vapour pressure of that component in pure state.