Question

Vapour pressure of water is 360 mm Hg, how much urea must be added to 200 ml water to reduce its vapour pressure by 0.5%? (molecular weight of urea =60)

Answer 1

Lowering of vapour pressure by a non-volatile solute and determination of molar masses from lowering vapour pressure are both colligative properties that is they depend upon only the number particles of solute present in the solution and not its identity.

Complete step by step solution:
We know that when a nonvolatile solute is dissolved in a liquid the vapour pressure of the solution becomes lower than the vapour pressure of the pure solvent. Let us therefore analyse that if dissolving n moles of urea in N moles of water. Then the mole fraction of the solvent(water), ${X_W} = \dfrac{N}{{N + n}}$ and mole fraction of urea is, ${X_U} = \dfrac{n}{{n + N}}$. Since urea is non-volatile, it would have negligible vapour pressure. The vapour pressure of the solution is merely the vapour pressure of the solvent (water). According to Raoult's law “ the vapour pressure of a solvent in the ideal solution is directly proportional to its mole fraction”.
${P_W} = {X_W}P_W^ \circ \\\ \because {X_W} + {X_U} = 1 \\\$
Where $P_W^ \circ$ is the vapour pressure of the pure solvent=360 mm Hg and then the equation can be written as,
${P_W} = (1 - {X_U})P_W^ \circ \\\ \Rightarrow \dfrac{{{P_W}}}{{P_W^ \circ }} = 1 - {X_U} \\\ \Rightarrow \dfrac{{P_W^ \circ - {P_W}}}{{P_W^ \circ }} = {X_U} \\\$
The expression on the left hand side is usually called the relative lowering of vapour pressure. Since mole fraction of urea is given by n/(n+N) or
$\dfrac{{\dfrac{{{w_u}}}{{{m_u}}}}}{{\dfrac{{{W_W}}}{{{M_W}}} + \dfrac{{{w_u}}}{{{m_u}}}}}$ ${P_W}$ is the reduced pressure given by ${P_w} = P_W^ \circ - 0.5P_W^ \circ = 0.5P_W^ \circ$ . Now substituting the values we get,
$\dfrac{{P_W^ \circ - {P_W}}}{{P_W^ \circ }} = \dfrac{{\dfrac{{{w_u}}}{{{m_u}}}}}{{\dfrac{{{W_W}}}{{{M_W}}} + \dfrac{{{w_u}}}{{{m_u}}}}}$
$
\Rightarrow \dfrac{{P_W^ \circ - 0.5P_W^ \circ }}{{{P_W}}}= \dfrac{{\dfrac{{{w_u}}}{{60}}}}{{\dfrac{{18}}{{200}} + \dfrac{{{w_u}}}{{60}}}} \\

\Rightarrow \dfrac{{360 - 358.2}}{{358.2}} = \dfrac{{\dfrac{{{w_u}}}{{60}}}}{{\dfrac{{200}}{{18}} + \dfrac{{{w_u}}}{{60}}}} \\
\Rightarrow 0.005 \times \dfrac{{200}}{{18}} \times 60 = {w_u} = 3.33g \\
$
Hence 3.33g of urea is required.

Note: The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute. Also raoult’s law is not applicable to solutes which dissociate or associate in a particular solution.