Question

A $0.0020M$ aqueous solution of an ionic compound $Co{\left( {N{H_3}} \right)_6}\left( {N{O_2}} \right)Cl$ freezes at $ - {0.00732^ \circ }C$. Number of moles of ions which $1$ mole of ionic compound produces on being dissolved in water will be: $\left[ {{k_f} = {{1.86}^ \circ }C/m} \right]$
A.$1$
B.$2$
C.$3$
D.$4$

Answer 1

Hint: This question gives the knowledge about depression in freezing point. Depression in freezing point is defined as the decrease in the freezing point of the solvents when solutes are added into the solution. It is a colligative property.
Formula used: The formula used to determine the depression in freezing point is as follows:
$\Delta {T_f} = i.{k_f}.m$
Where $\Delta {T_f}$ is the depression in freezing point, $m$ is the molality, ${k_f}$ is the cryoscopic constant and $i$ is the Van’t Hoff factor.
The formula used to determine the number of moles using degree of dissociation is as follows:
$\alpha = \dfrac{{i - 1}}{{n - 1}}$
Where $i$ is the Van’t Hoff factor, $\alpha $ is the degree of dissociation and $n$ is the number of moles.

Complete step by step answer:
Depression in freezing point is defined as the decrease in the freezing point of the solvents when solutes are added into the solution. It is a colligative property. The colligative properties generally depend upon the number of particles.
First we will determine the Van’t Hoff factor using the formula of depression in freezing point as follows:
$ \Rightarrow \Delta {T_f} = i.{k_f}.m$
Rearrange the formula to determine the Van’t Hoff factor as follows:
$ \Rightarrow i = \dfrac{{\Delta {T_f}}}{{{k_f}.m}}$
Substitute depression in freezing point as $0.0732K$, molality as $0.0020$ and cryoscopic constant as $1.86$ in the above formula.
$ \Rightarrow i = \dfrac{{0.0732}}{{\left( {1.86} \right) \times \left( {0.0020} \right)}}$
On simplifying, we get
$ \Rightarrow i = 2$
As we know, ionic compounds in the aqueous solutions dissociate completely. So, the degree of dissociation will be approximately equal to $1$.
Now we will determine the number of moles formed using degree of dissociation formula as follows:
$ \Rightarrow \alpha = \dfrac{{i - 1}}{{n - 1}}$
Substitute degree of dissociation as $1$ and Van’t Hoff factor as $2$ in the above formula.
$ \Rightarrow 1 = \dfrac{{2 - 1}}{{n - 1}}$
On simplifying, we get
$ \Rightarrow 1 = \dfrac{1}{{n - 1}}$
Rearrange the above formula to determine the number of moles:
$ \Rightarrow n - 1 = 1$
On simplifying, we get
$ \Rightarrow n = 1 + 1$
On further simplifying, we get
$ \Rightarrow n = 2$
Therefore, the number of mole ions formed are $2$.

Hence, option B is the correct option.

Note:
Always remember that the depression in freezing point is defined as the colligative property which experiences a decrease in the freezing point of the solvents when solutes are added into the solution. The colligative properties generally depend upon the number of particles.