Question

The van’t Hoff factor for 0.1M barium nitrate is 2.74. The percentage of dissociation of barium nitrate is:
(A) 91.3 %
(B) 87 %
(C) 100%
(D) 74 %

Answer 1

Hint: Vant’t hoff factor and degree of dissociation are related to each other as follows:$\alpha = \dfrac{{i - 1}}{{n - 1}}$.
Complete step by step answer: The van’t hoff factor is used to measure the extent of association or dissociation of molecules in the solution. It is denoted as ‘i’ and it is a constant value with no units.
The relation between Vant’t hoff factor(i) and degree of dissociation(α) can be depicted as follows:$\alpha = \dfrac{{i - 1}}{{n - 1}}$ (or) $i = 1 + \alpha (n - 1)$
where n is the number of particles in solution after dissociation.
It is given that the van’t Hoff factor for 0.1M barium nitrate is 2.74. We need to find the percentage of dissociation. First, we need to find the degree of dissociation.
For the given 0.1M barium nitrate solution the dissociation can be shown as below:
$Ba{(N{O_3})_2} \rightleftarrows B{a^{2 + }} + 2N{O_3}^{2 - }$
The number of particles after dissociation, n=3.
Substitute all the values in the formula to get the degree of dissociation,
$\alpha = \dfrac{{i - 1}}{{n - 1}} = \dfrac{{2.74 - 1}}{2} = \dfrac{{1.74}}{2} = 0.87 \\\ \Rightarrow \alpha = 0.87 \\\$
The degree of dissociation is 0.87 x 100= 87 %.
So, the correct option is B.
Additional Information: Van’t hoff factor can also be defined and calculated through following ways:
$i = \dfrac{{Normal{\text{ }}molecular{\text{ }}mass}}{{{\text{Observed }}molecular{\text{ }}mass}} \\\ i = \dfrac{{Observed{\text{ }}value{\text{ }}of{\text{ }}colligative{\text{ }}property}}{{Calculated{\text{ }}value\;of\;colligative{\text{ }}property}} \\\$
During association, the value of ‘i’ is less than one since molecular mass is more than the starting substance. During dissociation, the value of ‘i’ is greater than one since molecular mass is less than the starting substance. In case of no dissociation ‘i’ the value becomes equal to one.

Note: Another way to solve is to form the number of moles during and after dissociation.

{\text{ }}Ba{(N{O_3})_2} \rightleftarrows B{a^{2 + }} + 2N{O_3}^{2 - } \\\ At{\text{ t = 0 1 0 0}} \\\ At{\text{ }}equilibrium, \\\ {\text{ 1 - }}\alpha {\text{ }}\alpha {\text{ 2}}\alpha \\\ \end{gathered} $$ Total moles after dissociation= $0.1 - \alpha + \alpha + 2\alpha = 0.1 + 2\alpha $ $\begin{gathered} i = \dfrac{{Observed{\text{ }}value{\text{ }}of{\text{ }}colligative{\text{ }}property}}{{Calculated{\text{ }}value\;of\;colligative{\text{ }}property(before{\text{ }}association{\text{ }}or{\text{ }}dissociation)}} \\\ i = \dfrac{{1 + 2\alpha }}{1} \\\ 2.74 = \dfrac{{1 + 2\alpha }}{1} \\\ \Rightarrow \alpha = 0.87 \\\ \end{gathered} $ The degree of dissociation is 0.87 x 100= 87 %. So, the correct option is B.