Question

The vapour pressure of a dilute aqueous solution of glucose is $750mm{\text{ Hg}}$ at $373K$. Calculate (i) molality, (ii) mole fraction of the solute.
A.$0.73,0.0132$
B.$0.078,0.02$
C.$0.0729,0.0187$
D.$0.0779,0.3$

Answer 1

Colligative properties: It is defined as the ratio of number of solute particles to the number of solvent particles. It only depends on the number of ions and not on the nature of the solute.
Raoult’s law: It states that vapour pressure of an ideal solution is proportional to the mole fraction of the solvent.

Complete step by step answer:
Colligative properties: It is defined as the ratio of number of solute particles to the number of solvent particles. It only depends on the number of ions and not on the nature of the solute.
There are four colligative properties: Depression in freezing point, elevation in boiling point, osmotic pressure and lowering in vapour pressure.
Depression in freezing point: It is defined as the decrease in freezing point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Elevation in boiling point: It is defined as an increase in boiling point of a solvent in addition to non-volatile solute. It is directly proportional to the molality of the solution.
Osmotic pressure: It is defined as the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.
Relative lowering in vapour pressure: When non- volatile solutes are added to a solution then there is a decrease in its vapour pressure, which is known as relative lowering in vapour pressure.
Raoult’s law: It states that vapour pressure of an ideal solution is proportional to the mole fraction of the solvent.
One form of Raoult’s law is as: $\dfrac{{{{\text{P}}_ \circ }{\text{ - }}{{\text{P}}_{\text{s}}}}}{{{{\text{P}}_ \circ }}}{\text{ = }}\dfrac{{{{w \times M}}}}{{{{m \times W}}}}$ where ${P_ \circ }$ is the vapour pressure of pure solvent and ${P_s}$ is the vapour pressure of solution. $w$ is the mass of the solvent, $W$ is molar mass of the solvent, $m$ is mass of solute and $M$ is molar mass of the solute.
We know that ${P_ \circ }$ for water at $373K$ is $760mm{\text{ Hg}}$, ${P_s}$ is given as $750mm{\text{ Hg}}$ and value of $M$ is $18$. So,
$\dfrac{{760 - 750}}{{760}} = \dfrac{w}{{W \times m}} \times 18 \\\ \dfrac{w}{{W \times m}} = \dfrac{{10}}{{760 \times 18}} \\\$
We know that molality is defined as the moles of solute per kilograms of solvent.
So, molality $= \dfrac{w}{{W \times m}} \times 1000$ $= \dfrac{{10}}{{760 \times 18}} \times 1000 = 0.73m$
Also we know that vapour pressure of solution is equal to the product of mole fraction of solvent with vapour pressure of pure solvent.
${P_s} =$ mole fraction of solvent $\times {P_ \circ }$
Mole fraction of solvent $= \dfrac{{{P_s}}}{{{P_ \circ }}}$ $= \dfrac{{750}}{{760}} = 0.986$
So the mole fraction of solute will be $1 - 0.986 = 0.014$.

So option A is correct.

Note:
Mole fraction is defined as the ratio of number of moles of substance to the total moles. The sum of the mole fraction of solute and the mole fraction of solvent is one. So if we know the mole fraction of any one of them then we can calculate the mole fraction of others.