Question

Relative decrease in vapour pressure of an aqueous solution containing 2 moles $[Cu{(N{H_3})_3}Cl]Cl$ in 3 moles of ${H_2}O$ is 0.50. On reaction with $AgN{O_3}$ , this solution will form:
[assuming no change in degree of ionization of a 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

To solve this question, we must first calculate the mole fraction of the solute and then equate it with the lowering of the vapour pressure. After that we must find the Van’t Hoff factor and the degree of dissociation. Finally, we must write the chemical reaction and then calculate the number of corresponding chlorine ions, which would give a direct relation to the amount of AgCl formed.

Complete Step-by-Step Answer:

Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Mole fraction can be understood as the reaction of the total number of moles of a substance present in a compound, to the total number of moles of all the constituents of the compound.
Also, another important thing to keep in mind is that when impurities are introduced to pure solutions, they tend to alter the physical properties of the solution like boiling point, freezing point, vapor pressure, etc. Due to impurities, there is an elevation in the boiling point, a depression in the freezing point of the solution and a decrease in vapour pressure. While calculating these values, many a times, the compound does not dissociate into smaller ions. Hence, we consider the general case where we consider the solution to be one single unit. But there are ionic solutions which involve the dissociation into ions. Here, each ion must be accounted for because it too contributes to the variation of these physical properties. To rectify this situation, we use a correction factor known as the Van’t Hoff factor.
Now, moving back to the question,
The mole fraction of $[Cu{(N{H_3})_3}Cl]Cl$ $= \dfrac{{i.{n_1}}}{{i.{n_1} + {n_2}}}$ , where ${n_1}$ is the number of moles of solute, ${n_2}$ is the number of moles of solvent and i is the Van’t Hoff factor.
The mole fraction of $[Cu{(N{H_3})_3}Cl]Cl$ $= \dfrac{{i.2}}{{i.2 + 3}}$
The relative lowering in vapour pressure is:
$\dfrac{{\Delta P}}{P}$ = 0.5 = mole fraction
Hence, 0.5 = $\dfrac{{i.2}}{{i.2 + 3}}$
$\dfrac{1}{2} = \dfrac{{i.2}}{{i.2 + 3}}$
i = 1.5
Now, Van’t Hoff factor can be calculated using the formula
$i = 1 + \left( {n - 1} \right)\alpha$
$1.5 = 1 + \left( {2 - 1} \right)\alpha$
$\alpha$ = 0.5

The reaction of the given solution with $AgN{O_3}$ can be given as:
$[Cu{(N{H_3})_3}Cl]Cl + AgN{O_3} \to [Cu{(N{H_3})_3}Cl]N{O_3} + AgCl$
Hence the number of moles of $Cl{}^ -$ ions = 2 (0.5) = 1 mole
Hence, 1 mole of AgCl is precipitated.

Hence, Option A is the correct option

Note: Van’t Hoff factor can be understood as the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.