Question

$M{X_2}$ dissociates in ${M^{2 + }}$ and ${X^ - }$ ions in an aqueous solution with a degree of dissociation 0.5 ($\alpha $). The ratio of the observed depression of the freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is _______.

Answer 1

Hint: Basically, in this question van’t Hoff factor ($i$) is asked indirectly because van’t Hoff factor is the ratio of the observed colligative property to the calculated colligative property. Calculate the total observed moles for the dissociation reaction of $M{X_2}$. To find the value of $i$, take the ratio of total observed moles after dissociation to the initial moles before dissociation.

Complete step by step answer:
We are given that $M{X_2}$ dissociates in ${M^{2 + }}$ and ${X^ - }$ ions. Therefore, equation for dissociation reaction of $M{X_2}$is as follows:
$M{X_2} \rightleftharpoons {M^{2 + }} + 2{X^ - }$
Initial moles of $M{X_2}$ at the starting of the reaction will be 1, and moles of ${M^{2 + }}$ and ${X^ - }$ will be zero. If $\alpha $ is the degree of dissociation, then we would have $1 - \alpha $ moles of undissociated $M{X_2}$, $\alpha $ moles of ${M^{2 + }}$ and $\alpha $ moles of ${X^ - }$. Thus,
\[\begin{align}
  & \begin{matrix}
   \text{Reaction } & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M{{X}_{2}}\rightleftarrows & {{M}^{2+}}+ & 2{{X}^{-}} \\
\end{matrix} \\
 & \begin{matrix}
   \text{Initial moles} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ & 0\ \ \ \ \ \ \ \ & 0 \\
\end{matrix} \\
 & \begin{matrix}
   \text{Moles at equilibrium} & \ \text{ }1-\alpha \ \ \ \ & \alpha \ \ \ \ \ \ \ & 2\alpha \\
\end{matrix} \\
\end{align}\]

Therefore, total observed moles at equilibrium are: $1 - \alpha + \alpha + 2\alpha = 1 + 2\alpha $
Since,$\alpha $= 0.5 given.
$ \Rightarrow 1 + 2\alpha = 1 + 2(0.5) = 2$ moles.
Thus, total number of moles after dissociation = 2

 And, number of moles before dissociation of $M{X_2}$ = 1
Now, we know that van’t Hoff factor ($i$) is defined in terms of moles as:
$i = \dfrac{{{\text{Total number of moles after dissociation/association}}}}{{{\text{Number of moles before dissociation/association}}}}$

Therefore, $i$ for the dissociation reaction of $M{X_2}$ is:
$i = \dfrac{{{\text{Total number of moles after dissociation of M}}{{\text{X}}_2}}}{{{\text{Number of moles before dissociation of M}}{{\text{X}}_2}}}$
$ \Rightarrow i = \dfrac{2}{1} = 2$

$i$ is also defined as the ratio of observed colligative property to the calculated colligative property. We are asked the ratio of the observed depression of the freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation, which is actually $i$ because depression in freezing point is a colligative property. Hence, the required ratio is 2.

Note: Colligative properties are all those properties of solutions which depend only on the number of solute particles and are independent of their chemical identity. Depression in freezing point is among the four colligative properties of solutions.