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(D) For the dissociation of $K_4[Fe(CN)_6]$: $K_4[Fe(CN)_6] \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$ Here,the number of ions produced per formula un

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$\left(\frac{m}{1+4x}\right)$

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(D) For the dissociation of $K_4[Fe(CN)_6]$: $K_4[Fe(CN)_6] \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$ Here,the number of ions produced per formula unit is $n = 5$. Using the formula for the van't Hoff factor $(i)$ in terms of degree of dissociation $(x)$: $i = 1 + (n - 1)x = 1 + (5 - 1)x = 1 + 4x$. We know that the van't Hoff factor is also defined as the ratio of normal molecular weight to abnormal molecular weight: $i = \frac{m_{normal}}{m_{abnormal}}$. Therefore,$m_{abnormal} = \frac{m_{normal}}{i} = \frac{m}{1 + 4x}$.

If $m$ is the normal molecular weight and $x$ is the degree of dissociation of $K_4[Fe(CN)_6]$,then the abnormal molecular weight in the solution of the salt will be $:-$

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