Question

Relative decrease in vapour pressure of an aqueous solution containing $2$ moles $\left[ Cu{{\left( N{{H}_{3}} \right)}_{3}}Cl \right]Cl$ in $3$ moles ${{H}_{2}}O$ is $0.50$. On reaction with $AgN{{O}_{3}}$, this solution will form: [assuming no change in degree of ionization of substance on adding $AgN{{O}_{3}}$]
A. $1\,mol\,AgCl$
B. $0.25\,mol\,AgCl$
C. $0.5\,mol\,AgCl$
D. $0.40\,mol\,AgCl$

Answer 1

Hint: Lowering of vapour pressure is a colligative property. Colligative properties depend upon the concentration of the solute molecules.
The question can be solved using Raoult's law which states that partial vapour pressure is the product of partial vapour pressure of a pure solvent and mole fraction.

Formula used: $\dfrac{\Delta P}{P}=\dfrac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$
where, $P$ is the pressure, $\Delta P$ is the change in pressure, \[{{n}_{1}}\] is the number of solute and ${{n}_{2}}$ is the number of solvent.

Complete step by step answer:
When ionization of this complex compound takes place then the dissociation of the complex compound takes place.
$\left[ Cu{{\left( N{{H}_{3}} \right)}_{3}}Cl \right]Cl\to {{\left[ Cu{{\left( N{{H}_{3}} \right)}_{3}}Cl \right]}^{+}}+C{{l}^{-}}$
Here, the degree of ionization is $\alpha $.
$i=1+\alpha $
where, $i$ is the Van't Hoff factor and $\alpha $ is the degree of ionization.
When lowering of vapour pressure takes place, then the formula used is:
 $\dfrac{\Delta P}{P}=\dfrac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$
where, $P$ is the pressure, $\Delta P$ is the change in pressure, \[{{n}_{1}}\] is the number of solute and ${{n}_{2}}$ is the number of solvent.
Here, complex compounds are used, therefore, degree of ionization should be included.
Hence,
$\dfrac{\Delta P}{P}=\dfrac{{{n}_{1}}(1+\alpha )}{{{n}_{1}}(1+\alpha )+{{n}_{2}}}$
Substituting the values in above formula, we get,
$\dfrac{\Delta P}{P}=\dfrac{2(1+\alpha )}{2(1+\alpha )+3}$
$\Rightarrow \dfrac{1}{2}=\dfrac{2(1+\alpha )}{2(1+\alpha )+3}$
On solving further, we get,
$\Rightarrow \alpha =0.5$
Total $C{{l}^{-}}$ ions produced: $2\alpha \Rightarrow 2\times 0.5=1$
Number of moles of $AgCl$ produced is $1\,mol$.

So, the correct answer is Option A.

Additional information:
Colligative properties are defined as the properties that depend on the concentration of solute molecules. Colligative properties show changes in boiling point elevation, freezing point depression, osmotic pressure and lowering of vapour pressure.
The vapour pressure of pure solvent is greater than the vapour pressure of a solution containing non – volatile liquid.
Adding a solute decreases the vapour pressure because the additional solute particles fill the gaps between the solvent particles and cover up the space.
Raoult’s law states that the partial vapour pressure of a molecule is the product of vapour pressure of pure solvent and mole fraction of a molecule.
$P=P{}^\circ {{x}_{A}}$
where, $P$ is the partial pressure, $P{}^\circ $ is the partial vapour pressure of pure solvent and ${{x}_{A}}$ is the mole fraction.

Note: When a solute is added, vapour pressure decreases.
To calculate the number of moles of $AgCl$ , you should know how to calculate degree of ionization using lowering of vapour pressure formula.
The degree of ionisation is defined as the proportion of neutral particles that are ionised to charged particles.