If the Van't Hoff factor for $0.1{\text{M}}$ ${\text{Ba}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ solution is $2.74$, the degree of dissociation is:
A.$0.87$
B.$0.74$
C.$0.91$
D.$87$

Hint: The degree of dissociation may be defined as the fraction of the total substance that undergoes dissociation, i.e., degree of dissociation is equal to the number of moles dissociated divided by the total number of moles taken.
In general, the dissociation of an electrolyte A into ‘n’ ions is shown below.
${{\text{A}}_{\text{n}}} \to {\text{nA}}$
If ‘i’ is the van’t Hoff factor, then for this dissociation, the degree of dissociation ${{\alpha }}$ is given by
${{\alpha = }}\dfrac{{{\text{i - 1}}}}{{{\text{n - 1}}}}$

Complete step by step answer:
For the general dissociation reaction
${{\text{A}}_{\text{n}}} \to {\text{nA}}$
If we start with 1 mole of the solute at equilibrium, then we will have $1 - {{\alpha }}$ moles of undissociated molecules and ${{n\alpha }}$ moles of the ions. Hence, the total number of moles of ions and the undissociated solute molecules will be equal to $1 - {{\alpha + n\alpha }} = 1 + \left( {{\text{n}} - 1} \right){{\alpha }}$.
Thus, the observed value of colligative property $ \propto 1 + \left( {{\text{n}} - 1} \right){{\alpha }}$
And since 1 mole of A was taken, theoretical value of colligative property $ \propto 1$
Since van’t Hoff factor is equal to the ratio of the observed value of colligative property to the theoretical value of the colligative property, therefore,
${\text{i}} = \dfrac{{1 + \left( {{\text{n - 1}}} \right){{\alpha }}}}{1}$
Or ${{\alpha = }}\dfrac{{{\text{i - 1}}}}{{{\text{n - 1}}}}$
Now, for the given electrolyte ${\text{Ba}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$, the dissociation will be
$\mathop {{\text{Ba}}{{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)}_{\text{2}}}}\limits_{1 - {{\alpha }}} \to \mathop {{\text{B}}{{\text{a}}^{2 + }}}\limits_{{\alpha }} + \mathop {2{\text{N}}{{\text{O}}_{\text{3}}}^ - }\limits_{2{{\alpha }}} $
Since n is equal to 3, we have
${{\alpha = }}\dfrac{{{\text{i - 1}}}}{{{\text{3 - 1}}}}$
According to the question, the van’t Hoff factor ‘i’ is equal to $2.74$.
Thus, the degree of dissociation for $0.1{\text{M}}$ ${\text{Ba}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ solution is
$
  {{\alpha = }}\dfrac{{{\text{2}}{\text{.74 - 1}}}}{{{\text{3 - 1}}}} \\
   \Rightarrow {{\alpha = }}\dfrac{{1.74}}{2} \\
   \Rightarrow {{\alpha = 0}}{\text{.87}} \\
 $

So, the correct option is A.

Note:
Colligative property is found to be inversely proportional to the molecular mass of the solute and so the van’t Hoff factor ‘i’ can also be defined as the ratio of the calculated molecular mass to the experimental molecular mass.If ‘i’ is greater than 1, there is dissociation of the solute in the solution and if ‘i’ is less than 1, there is association of the solute in the solution. For hundred percent dissociation of a solute in the solution, the van’t Hoff factor ‘i’ will be equal to the number of ions produced from one molecule of the solute.