Question

Vapour pressure of pure water at 298K is 23.8mm Hg.50g of urea \[N{{H}_{2}}CON{{H}_{2}}\] is dissolved in 850g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Answer 1

Hint: Vapour pressure of pure water will be lowered when a solute is added. As more solute is added, vapour pressure goes on lowering. Relative lowering of vapour pressure is ratio of lowering of vapour pressure to vapour pressure of pure component.

Complete answer:
Raoult’s Law states that partial vapour pressure of each component is directly proportional to its mole fraction present in the component.
\[\dfrac{23.8-p}{23.8}=\dfrac{0.83}{0.83+0.021}\]
${{P}_{1}}^{0}$is vapour pressure of pure component 1
\[\begin{align}
  & {{P}_{2}}\propto {{\text{x}}_{2}} \\
 & ={{P}_{2}}={{P}_{2}}^{0}{{x}_{2}} \\
\end{align}\]
${{P}_{2}}^{0}$is vapour pressure of pure component 2
\[\begin{align}
  & {{P}_{T}}={{P}_{1}}+{{P}_{2}} \\
 & ={{P}_{1}}^{0}.{{x}_{1}}+{{P}_{2}}^{0}.{{x}_{2}} \\
 & ={{P}_{1}}^{0}.{{x}_{1}}+{{P}_{2}}^{0}(1-{{x}_{1}}) \\
 & ={{P}_{1}}^{0}.{{x}_{1}}+{{P}_{2}}^{0}-{{P}_{2}}^{0}.{{x}_{1}} \\
 & \\
\end{align}\]
As Solute is nonvolatile, it will not exert any vapour pressure, so vapour pressure of component 2 is zero, hence
${{P}_{1}}={{P}_{1}}^{0}.{{x}_{1}}$
Lowering of vapour pressure is the difference between vapour pressure of pure components and vapour pressure of solution.
\[\Delta \text{P= }{{\text{P}}_{1}}^{0}-P\]
$={{P}_{1}}^{0}-{{P}_{1}}^{0}.{{X}_{1}}={{P}_{1}}^{0}(1-{{x}_{2}})={{P}_{1}}^{0}.{{x}_{2}}$
Relative lowering of vapour pressure is ratio of lowering of vapour pressure to vapour pressure of pure component.
Relative lowering of vapour pressure=$\dfrac{{{P}_{1}}^{0}.{{x}_{2}}}{{{P}_{1}}^{0}}={{x}_{2}}$
\[{{x}_{2}}\] is mole fraction of solution which is ratio of number of moles of solute to sum of moles of all components present in solution.
\[{{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}\]
\[{{n}_{2}}=\dfrac{50}{60}\]
\[{{n}_{1}}=\dfrac{850}{18}\]
So, \[\dfrac{23.8-p}{23.8}=\dfrac{0.83}{0.83+2.77}\]
So vapour pressure of water for this solution is 23.4 mm of Hg.
Relative lowering is 0.0173

Note:
Relative lowering of vapour pressure can be determined by calculating mole fraction of solute. When mole of solute is very small compared to moles of solvent, mole of solute can be neglected. Lowering of vapour pressure can be calculated by taking the difference between vapour pressure of pure components and solution.