Question

The vapour pressure of a dilute aqueous solution of glucose is $750{\text{ mm}}$$Hg$ at $373K$ calculate the molality and mole fraction of solution.

Answer 1

Hint:
Molality is defined as the number of moles of solute present per kg of solvent. The mole fraction is defined as the ratio of mole of substance to the total moles present in the solution.

Complete step by step answer:
As we know that the vapour pressure is defined as the pressure exerted by vapours at a particular temperature. We also know that vapour pressure of pure solution at $373K$ is $760nm{\text{ }}Hg$ where in this question the vapour pressure at $373K{\text{ }}\operatorname{is} {\text{ 750mmHg}}$. This means that a non-volatile solute (Glucose) is added to the solution. Now according to Raoult’s Law, we know that vapour pressure of the solution is directly proportional to the mole fraction of the solvent because solute is non-volatile.
i.e. ${P_s} \propto {x_1}$
Or ${P_s} = P^\circ {x_1}$
Or $\dfrac{{{P_s}}}{{P^\circ }} = {x_1}{\text{ - - - - - - - - (1)}}$
Here ${P_s}$ be the vapour pressure of solution and $P^\circ $ be the vapour pressure of solution and ${x_1}$ be the mole fraction of solvent.
$\therefore $ mole fraction of \[ = {x_2} = 1 - {x_1}\]
Solute
$\therefore {x_2} = 1.\dfrac{{{P_s}}}{{P^\circ }}$ [Putting values from equation $(1)$]
OR ${x_2} = \dfrac{{P^\circ - {P_s}}}{{P^\circ }}{\text{ - - - - - - }}\left( 2 \right)$
Now we know that vapour pressure of pure solvent is $760mmHg\left( {P^\circ } \right)$ and vapour pressure of solution given is$750mmHg\left( {{P_s}} \right)$. Putting values in equation $(2)$, we get
${x_2} = \dfrac{{760mmHg - 750mmHg}}{{760mmHg}} = \dfrac{{10}}{{760}}$
$\therefore {x_2} = \dfrac{{10}}{{760}}{\text{ - - - - - - }}\left( 3 \right)$
OR ${x_2} = 0.013$ Which is the required mole fraction of solution.
Now molality is defined as the moles of substance present per kg of solvent. If mole fraction of solute is given, it can be calculated using the formula.
$m = \dfrac{{1000.{x_2}}}{{\left( {1 - {x_2}} \right){m_1}}}{\text{ - - - - - - - - - }}\left( 4 \right)$
Here ${m_1}$ is the molecular mass of solvent. Here according to the question, aqueous solution of glucose is formed means that solute is glucose and solvent is water $\left( {{H_2}O} \right)$. The molecular weight of water as we all know is $18g$ and the value of ${x_2}$ is already evaluated in Equation $(3)$.
Putting all values in equation $(4)$, we get
$m = \dfrac{{1000 \times \dfrac{{10}}{{760}}}}{{\left( {\dfrac{{1 - 10}}{{760}}} \right) \times 18}} = \dfrac{{\dfrac{{10000}}{{760}}}}{{\dfrac{{750 \times 18}}{{760}}}} = \dfrac{{10000}}{{13500}} = 0.74m$
Hence required molality of the solution is $0.73$ mole and mole fraction of solute is $0.013$.

Note: The term $\dfrac{{P^\circ - {P_s}}}{{P^\circ }}$ is defined as the relative lowering in vapour pressure and is a colligative property as it depends on the number of moles of the substance. The relative lowering in vapour pressure is the ratio of vapour pressure lowering of solvent from solution to the vapour pressure of pure solvent.