Question

The vapour pressure of water at ${{20}^{\circ }}\,C$ is 17 mm Hg. Calculate the vapour pressure of a solution containing 2.8 g of urea $(N{{H}_{2}}CON{{H}_{2}})$ in 50 g of water N = 14, C = 12, O = 16, H = 1 (16.71 mm Hg).

Answer 1

Hint: The vapour pressure of the solution is dependent on the solute molecules dissolved in it. This vapour pressure is always lower than the vapour pressure of the pure solvent.

Complete step by step solution:
It is given that a solution is formed by the dissolving urea in water.
Given: Weight of water taken $({{m}_{1}})=50g$
Weight of urea dissolved $({{m}_{2}})=2.8g$
Then, the molecular mass of urea,$(N{{H}_{2}}CON{{H}_{2}})$ will be ${{M}_{2}}=(2\times N)+(4\times H)+C+O$
$=(2\times 14)+(4\times 1)+12+16=60g/mol$
and the molecular mass of water is ${{M}_{1}}=18g/mol$.
So, the moles of urea will be $\text{n=}\dfrac{\text{mass of urea}}{\text{molecular mass}\,\text{of urea}}=\dfrac{2.8}{60}=0.047\text{moles}$
and the moles of water will be $N=\dfrac{50}{18}=2.78\,\text{moles}$
the mole fraction of urea will be${{\chi }_{2}}=\dfrac{n}{n+N}=\dfrac{0.047}{0.047+2.78}=0.0166$ -----(a)
Also, vapour pressure of water is given to be ${{\text{P}}_{\text{1}}}^{\circ }\text{=17 mm}\,\text{Hg}$. -----(b)
Due to addition of urea in water, the vapour pressure of the solvent decreases. Thus, the vapour pressure of the solution $(P)$ is less than the pure solvent. This is known as the lowering in vapour pressure, given as $\Delta P={{P}^{\circ }}_{1}-P$ ------(c)
Then, the relative lowering of vapour pressure which is a colligative property, depends on the mole fraction of the urea, ${{\chi }_{2}}$ (solute) in solution, which is given as follows:
\[\dfrac{\Delta P}{P_{1}^{\circ }}={{\chi }_{2}}\]
Substituting the values of (a), (b) and (c) in the above equation, we get,
\[\dfrac{P_{1}^{\circ }-P}{P_{1}^{\circ }}=0.0166\]
\[\dfrac{17-P}{17}=0.0166\]

Therefore, vapor pressure of the solution $P=16.71\,mm\,Hg$.

Note: We have seen a lowering in vapour solution ($P=16.71\,mm\,Hg$) compared to the vapour pressure of water (${{\text{P}}_{\text{1}}}^{\circ }\text{=17 mm}\,\text{Hg}$). This is because the urea molecules being non-volatile in nature, do not contribute to the vapour pressure. Rather, it hinders the water molecules from escaping the surface of the solution. Thus, lowering its vapour pressure.