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**Hint:**Van’t Hoff factor is denoted by I and it is a measure of effect of solute on colligative properties such as osmotic pressure, relative lowering in vapour pressure, boiling point elevation and freezing point depression. We can evaluate the degree of dissociation by equating the dissociated species and finding $\dfrac{i-1}{n-1}$

**Complete answer:**The given salt in the question is MX 2

Van T Hoff factor (i) = 2

Steps: We know that the electrolyte MX 2 dissociates into M and 2 X in aqueous solution.

The reaction is shown below,

$M{{X}_{2}}\,\,\to \,\,M+2X$

We see that initially there is one mole of $M{{X}_{2}}$. Suppose $\alpha $ moles of $M{{X}_{2}}$ dissociates in the aqueous solution. Now after the dissociation, at equilibrium condition the reaction has $\alpha $ moles of M, $2\alpha $ moles of 2 X produced and (1- $\alpha $) moles of $M{{X}_{2}}$ left. where alpha is the degree of dissociation.

$M{{X}_{2}}\,\,\to \,\,M+2X$

$1-\alpha \,\,\,\,\,\propto \,\,\,2\alpha $

Therefore, total moles after the dissociation are given by $(1-\alpha )+\alpha +2\alpha $

$\Rightarrow \,1+2\alpha $ moles

Therefore, there are $1+2\alpha $ moles after the dissociation.

Also, we know that total moles before dissociation was one.

Now Van't Hoff factor $\left( i \right)$ is equal to the total moles after disassociation to initial moles,

$i=\dfrac{\text{Total moles after dissociation}}{\text{initial moles}}$

$\therefore \,\,i=\dfrac{1+2\alpha }{1}\,=\,1+2\alpha $

Now, we know that alpha ($\alpha $) is given by the following formula,

$\alpha =\dfrac{i-1}{n-1}$

Where, n is the number of species formed after disassociation.

It is already given in the question that $i$ is $2$

Also, n is the number of species formed after disassociation. And it clearly seen from the above dissociation equation that after dissociation total $3$ moles ( $2$moles of$X$and $1$mole of$M$) are formed.

Hence, we can say that n=$3$

So, putting the value of $i$ and $n$ in the above formula we get,

$\alpha =\dfrac{2-1}{3-1}=\dfrac{1}{2}$

\[\Rightarrow \,\alpha =0.50\]

Hence, the degree of dissociation is $0.50$.

**So, the option A is the correct answer.**

**Note:**The Van't Hoff factor is essentially $1$ for most non-electrolytes dissolved in water. For disassociation the Van T Hoff factor is greater than $1$ while for association Van T Hoff factor is less than $1$. Students must solve similar problems to have a good idea.

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