Question

What will be the value of degree of dissociation of $0.1{\text{ M}}$ $Mg{\left( {N{O_3}} \right)_2}$ solution if van’t Hoff factor is
(i){\text{ 78% }}
(ii){\text{ 92% }}
(iii){\text{ 87% }}
(iv){\text{ 100% }}

Answer 1

For the given electrolyte we have to calculate the van’t Hoff factor. Van’t Hoff factor is the ratio of the actual concentration of particles formed when the given compound is dissociated and concentration of the compound as calculated by its mass. With the help of this ratio we can calculate the van’t Hoff factor.

Complete answer:
When the $Mg{\left( {N{O_3}} \right)_2}$ starts dissociating into its ions then we can write the dissociation equation as:
$Mg{\left( {N{O_3}} \right)_2} \to M{g^{ + 2}} + 2N{O_3}^ -$
When the dissociation of the given compounds starts, let the amount of $Mg{\left( {N{O_3}} \right)_2}$ be present is one mole. Let us consider at a certain time the dissociation starts and let the degree of dissociation be $\alpha$ . Therefore it can be written as:
$Mg{\left( {N{O_3}} \right)_2} \to M{g^{ + 2}} + 2N{O_3}^ -$

At , $t{\text{ = 0}}$	1	0	0
At, $t{\text{ = }}{{\text{t}}_ \circ }$	$1 - \alpha$	$\alpha$	$2\alpha$

Therefore the total number of moles which are present at $t{\text{ = }}{{\text{t}}_ \circ }$ are:
$= {\text{ }}1 - \alpha {\text{ + }}\alpha {\text{ + 2}}\alpha$
$= {\text{ }}1{\text{ + 2}}\alpha$
The total number of moles which are present is equal to the van't Hoff factor. It is denoted by $i$ . Hence it can be represented as:
$i{\text{ = 1 + 2}}\alpha$
From the above equation we can calculate the value of degree of dissociation as:
$\alpha {\text{ = }}\dfrac{{i{\text{ - 1}}}}{2}$
We are provided with the value of van’t Hoff factor which is equal to $2.74$ . On putting the values we get,
$\alpha {\text{ = }}\dfrac{{{\text{2}}{\text{.74 - 1}}}}{2}$
$\alpha {\text{ = }}\dfrac{{1.74}}{2}$
$\alpha {\text{ = 0}}{\text{.87}}$
The percentage of dissociation can be calculated by converting it into a percentage which is equal to $87\%$ .
Thus the correct option is (iii){\text{ 87% }} .

Note:
Hence the degree of dissociation is meant for only ionic compounds. For the non-electrolyte compound the value of van’t Hoff factor is always one. If the degree of dissociation is large then the reaction will take less time to complete. Initially we take one mole of reactant, as it is easy for further calculations. The dissociation takes place in equilibrium as rate of forward reaction and backward remains equal ideally.