For a given electrolyte A_xB_y,the relationsh | vedclass

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(A) For an electrolyte $A_xB_y$ dissociating as $A_xB_y \rightarrow xA^{y+} + yB^{x-}$,the total number of ions produced is $n = x + y$.If '$\alpha$'

Vedclass · Accepted Answer

$\alpha = \frac{i - 1}{x + y - 1}$

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(A) For an electrolyte $A_xB_y$ dissociating as $A_xB_y \rightarrow xA^{y+} + yB^{x-}$,the total number of ions produced is $n = x + y$.If '$\alpha$' is the degree of dissociation,the number of particles at equilibrium is $1 - \alpha + x\alpha + y\alpha = 1 + \alpha(x + y - 1)$.The van't Hoff factor '$i$' is defined as the ratio of the number of particles after dissociation to the number of particles before dissociation.Thus,$i = 1 + \alpha(x + y - 1)$.Rearranging for '$\alpha$',we get $\alpha = \frac{i - 1}{x + y - 1}$.

For a given electrolyte $A_xB_y$,the relationship between the degree of dissociation '$\alpha$' and the van't Hoff factor '$i$' is ........

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