Question

Vapour pressure of water is $12.3$ $kpa$ at $300$ K. Calculate Vapor pressure of 1$m$ solution of a non-volatile solute in it? ($12.083$ $kpa$)

Answer 1

Hint: Vapour pressure is a measure of the tendency of a material to change from liquid state to gaseous or vapour state and the vapour pressure increases with increase in the temperature. The vapour pressure can be expressed in Pascal ($pa$), kilopascals ($kpa$) or pounds of force per square ($PSI$). If a substance has high vapour pressure at normal temperature, then the substance is volatile in nature.

Complete answer:
The vapour pressure can be determined as follows:
As per the question,
Vapour pressure of water = $12.3$= \[{P^0}\]
Molality = 1
We need to find out,
Vapour pressure of solute = \[x\]
$1000g$ of water contains 1 mole of solute
Number of moles of water = $\dfrac{{1000}}{{18}}$ = $55.56$$moles$
Mole fraction of solute (\[X\]) = $\dfrac{{moles{\text{ }}of{\text{ }}solute}}{{total{\text{ }}moles}}$ =$\dfrac{1}{{1 + 55.56}}$= $0.0177$
By applying Raoult’s law, the relative lowering in the vapour pressure is equal to the mole fraction of the solute $\dfrac{{{P^{0 - }}P}}{{{P^0}}}$ = (\[X\])
$ \Rightarrow $$\dfrac{{12.3 - P}}{{12.3}}$ = $0.0177$
$ \therefore $$12.08$

The vapour pressure of the solution is $12.08$ $kpa$.

Note:
As the temperature of the liquid increases, the kinetic energy also increases, the number of molecules transitioning into vapour also increases, and hence resulting in the increase in the vapour pressure. If a non-volatile solute is dissolved into a solvent to form an ideal solution, the vapour pressure of the final solution will be lower than that of the solvent. Hence, the decrease in the vapour pressure is directly proportional to the mole fraction of solute in an ideal solution.