Question

The relative lowering of vapour of an aqueous solution containing a non-volatile solute is $0.0125$ . The molality of the solution is:
A.$0.70$
B.$0.30$
C.$0.125$
D.$0.07$

Answer 1

According to Raoult's law, relative lowering of vapour pressure of solution is equal to mole fraction of non-volatile solute in the solution. It is a colligative property and only depends on the quantity of the particles and not their nature.

Complete step by step answer:
Vapour pressure is defined as the pressure exerted by the gas on the surface of liquid in a closed vessel. It is a surface phenomenon. The molecules present at the surface have lesser force of attraction and have more kinetic energy than rest molecules of the liquid. Vapour pressure is defined for volatile liquids and vapour pressure of non-volatile liquids is taken to be zero. The more the volatility of the liquid is, more will be its vapour pressure. For a solution containing both solute and solvent of volatile nature, vapour pressure of that solution is the sum of vapour pressure of both solvent and solute. For these types of solution, the partial vapour pressure of each component is directly proportional to the mole fraction of that component. On the other hand, for a solution containing non-volatile solute, vapour pressure is equal to partial vapour pressure of volatile solvent. The vapour pressure of a solution containing non-volatile solute is less than vapour pressure of pure solvent because of a decrease in surface area for the volatile molecules. For such a solution, relative lowering of vapour pressure is equal to mole fraction of non-volatile solute.
As given in our question, relative lowering of vapour pressure of non-volatile solute is $0.0125$ . This means in this solution, mole fraction of non-volatile solute is 0.0125.
As mole fraction of solute is equal to moles of solute divided by total number of moles, solute plus solvent.
$\dfrac{n}{{\left( {{\text{n}} + {\text{N}}} \right)}} = 0.0125$
Here n is moles of solute and N is moles of solvent
We can say if n is $0.0125$ , then $\left( {{\text{n}} + {\text{N}}} \right) = 1$ and thereby
${\text{N}} = {\text{ }}1 - 0.0125 = 0.9875$
And the molar mass of solvent is 18 amu as solvent is water.
${\text{Molality }} = \dfrac{{{\text{moles}}}}{{{\text{volume in mL}}}} \times 1000$
${\text{Molality }} = 0.0125 \times 1000$
$\Rightarrow {\text{Molality }} = 0.9875 \times 18 = 0.7$

Thus, the correct option is A.

Note:
Lowering of vapour pressure is vapour pressure of pure solvent minus vapour pressure of solution. Relative lowering of vapour pressure is lowering of vapour pressure divided by vapour pressure of pure solvent.