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(A) The dissociation of the salt $MX_2$ is represented as: $MX_2 \rightleftharpoons M^{2+} + 2X^-$. Let the initial concentration be $1$ mole. At equi

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$0.50$

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(A) The dissociation of the salt $MX_2$ is represented as: $MX_2 \rightleftharpoons M^{2+} + 2X^-$. Let the initial concentration be $1$ mole. At equilibrium,the moles are: $M^{2+} = \alpha$,$X^- = 2\alpha$,and $MX_2 = 1 - \alpha$,where $\alpha$ is the degree of dissociation. The total number of particles at equilibrium is $(1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha$. The van't Hoff factor $i$ is given by the ratio of the total number of particles at equilibrium to the initial number of particles: $i = \frac{1 + 2\alpha}{1}$. Given $i = 2$,we have $1 + 2\alpha = 2$. Solving for $\alpha$: $2\alpha = 1$,which gives $\alpha = 0.50$.

An aqueous solution of a salt $MX_2$ at a certain temperature has a van't Hoff factor of $2$. The degree of dissociation for this solution of the salt is

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