Question

18.2g of urea is dissolved in 100g of water at ${{50}^{\circ }}C$. The lowering of vapour pressure produced if 5 mm Hg. Calculate the molecular mass of urea. The vapour pressure of water at ${{50}^{\circ }}C$ is 92 mm Hg.
(A)47.05
(B) 56.05
(C) 57.05
(D) 67.05

Answer 1

Hint: The relative lowering of vapour pressure is a colligative property, that is it depends on the number of solute particles. Addition of non volatile solute, lowers the vapour pressure of the solvent.


Complete answer:
Vapour pressure is the pressure exerted by the vapours over the liquid at a given temperature under the equilibrium conditions. This lowering in vapour pressure is because of the fact that after the addition of the solute to the solvent(pure solvent), the surface of the liquid now has molecules of both, the pure liquid and the solute. The number of solvent molecules escaping into vapour phase gets reduced , resulting in decrease in the pressure exerted by the vapour phase. This is known as relative lowering of vapour pressure. This lowering in vapour pressure depends on the concentration of non-volatile solute added in the solution independent of its nature and therefore, it is one of the colligative properties.
As we know the formula of relative lowering of vapour pressure is as follows,
     \[\dfrac{{{p}_{A}}{}^\circ -{{p}_{A}}}{{{p}_{A}}{}^\circ }={{x}_{2}}\] …(i)
Where,${{p}_{A}}{}^\circ $= vapour pressure of pure solvent,
Given value of ${{p}_{A}}{}^\circ $ =92 mm Hg
${{p}_{A}}$= vapour pressure of the solution,
${{p}_{A}}{}^\circ -{{p}_{A}}$ = relative lowering of vapour pressure,
Given value of ${{p}_{A}}{}^\circ -{{p}_{A}}$= 5 mm Hg
${{x}_{2}}$= mole fraction of the solute
Also, we know the formula of mole fraction is,
     \[{{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}\]
where, ${{n}_{2}}$= moles of solute,
${{n}_{1}}$= moles of solvent
Here, the solvent is water, and the molecular weight of water is 18, and the weight of water is 100g, also the weight of urea is also given as 18.2g.
Substituting the above values in the formula of mole fraction,
     \[\begin{align}
  & {{x}_{2}}=\dfrac{\dfrac{18.2}{{{M}_{2}}}}{\dfrac{100}{18}} \\
 & \Rightarrow {{x}_{2}}=\left( \dfrac{18.2}{{{M}_{2}}} \right)\left( \dfrac{18}{100} \right) \\
 & \Rightarrow {{x}_{2}}=\dfrac{3.26}{{{M}_{2}}} \\
\end{align}\]
here, ${{M}_{2}}$= molecular mass of urea${{M}_{2}}$
 From here, we get the value of ${{x}_{2}}$, Now we will substitute the value of ${{x}_{2}}$in equation (i),
     \[\dfrac{5}{92}=\dfrac{3.26}{{{M}_{2}}}\]
     \[\begin{align}
  & {{M}_{2}}=\dfrac{3.26\times 92}{5} \\
 & {{M}_{2}}=59.9 \\
\end{align}\]
So we get the molecular weight of urea is 59.9g/mol.

So, the correct answer is “Option C”.

Note: Colligative properties are the physical changes that results from addition of solute to solvent. It does not depend on the nature of solute, however, in relative lowering of vapour pressure the solute must be non volatile.