The relationship between the van’t Hoff facto | vedclass

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(D) The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by: $\alpha = \frac{i - 1}{n' - 1}$ Where,

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$\alpha = \frac{i - 1}{n' - 1}$

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(D) The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by: $\alpha = \frac{i - 1}{n' - 1}$ Where,$n'$ is the number of ions formed after dissociation. For the dissociation reaction: $A \rightleftharpoons n'B$ Initially: $1$ mole,$0$ After dissociation: $(1 - \alpha)$ mole,$n'\alpha$ Total number of moles present in the solution: $= (1 - \alpha) + n'\alpha = 1 + (n' - 1)\alpha$ By definition,the van’t Hoff factor $(i)$ is the ratio of observed colligative property to calculated colligative property,which equals the ratio of total moles after dissociation to initial moles: $i = \frac{1 + (n' - 1)\alpha}{1} = 1 + (n' - 1)\alpha$ Rearranging for $\alpha$: $i - 1 = (n' - 1)\alpha$ $\alpha = \frac{i - 1}{n' - 1}$

The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is:

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