Question

The vapour pressure of water is $92mm$ at $323K$. $18.1g$ of urea are dissolved in $100g$ of water. The vapour pressure is reduced by $5mm$. Calculate the molar mass of urea.

Answer 1

The relative lowering of vapour pressure is directly proportional to the mole fraction of the solute present in the solution. Thus, as we are given the lowering value, we can calculate the mole fraction. Since, the weight of the solute added is also given, we can find the molar mass of urea by dividing the given mass with the mole fraction.
Formulas used: $\dfrac{{p_A^0 - {p_A}}}{{p_A^0}} = {x_B}$
Where $p_A^0$ and ${p_A}$ denotes the vapour pressures of the pure solvent and lowered vapour pressure respectively, and ${x_B}$ denotes the mole fraction of the solute.
${x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$ where ${n_A}$ and ${n_B}$ denote the number of moles of solvent and solute
Number of moles of a component $= \dfrac{W}{M}$ where $W$ is the given mass and $M$ is the molar mass.

Complete step by step answer:
The relative lowering of vapour pressure when a solute is added to a solvent can be found by the formula:
$\dfrac{{p_A^0 - {p_A}}}{{p_A^0}} = {x_B}$ ……………………….. (1)
Where $p_A^0$ and ${p_A}$ denotes the vapour pressures of the pure solvent and lowered vapour pressure respectively, and ${x_B}$ denotes the mole fraction of the solute.
As we know, the mole fraction of solute in a solution is given by:
${x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}$ where ${n_A}$ and ${n_B}$ denote the number of moles of solvent and solute ……… (2)
Number of moles of a component $= \dfrac{W}{M}$ where $W$ is the given mass and $M$ is the molar mass.
Given mass of water is $100g$ and molar mass of water is $(1 \times 2) + (16) = 18g$. Hence,
Number of moles of water, ${n_A} = \dfrac{{100}}{{18}} = 5.56$
Given mass of urea is $18.1g$ and we have to find its molar mass (${M_B}$). Hence,
Number of moles of urea, ${n_B} = \dfrac{{18.1}}{{{M_B}}}$
Substituting the values of ${n_A}$ and ${n_B}$ in equation (2), we get:
${x_B} = \dfrac{{\dfrac{{18.1}}{{{M_B}}}}}{{5.56 + \dfrac{{18.1}}{{{M_B}}}}}$
Substituting this value, and the values of lowering of vapour pressure, $p_A^0 - {p_A} = 5mm$ and vapour pressure of pure solvent, $p_A^0 = 92mm$ in equation (1), we get:
$\dfrac{5}{{92}} = \dfrac{{\dfrac{{18.1}}{{{M_B}}}}}{{5.56 + \dfrac{{18.1}}{{{M_B}}}}}$
Rearranging this, we get:
$0.054\left( {5.56 + \dfrac{{18.1}}{{{M_B}}}} \right) = \dfrac{{18.1}}{{{M_B}}}$
$\Rightarrow 0.3 + \dfrac{{0.97}}{{{M_B}}} = \dfrac{{18.1}}{{{M_B}}}$
On solving further, we get:
$0.3 = \dfrac{{17.73}}{{{M_B}}} \Rightarrow {M_B} = \dfrac{{17.73}}{{0.3}}$
On solving this, we get: ${M_B} = 57.1g$
Hence, the molar mass of urea is $57.1g/mol$

Note: Vapour pressure is the pressure exerted by the vapours of a liquid above its surface at equilibrium. When a non-volatile solute is added to the solvent, it acts as an obstruction for the solvent molecules to go into the vapour phase, and thus, causes a lowering of the vapour pressure. Note that relative lowering of vapour pressure is a colligative property. That is, it is entirely dependent on the amount of solute added and not its nature.