What will be the equivalent weight of $C u S O_{4}$ in terms of its molecular weight m in the following reaction?
$C u S O_{4} + 4 K I \to C u_{2} I_{2} + I_{2} + K_{2} S O_{4}$

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Solution

$+ 2 2 C u S O_{4} 2 m o l e s + 4 K I ⟶ + 1 C u_{2} I_{2} 1 m o l e + I_{2} + K_{2} S O_{4}$
$∴ n - f a c t o r = 1 / 2 = 0.5$
The equivalent weight of $C u = \frac{Normal weight}{n-factor} = \frac{63.5}{0.5} = 127 u$
Mass of $S O_{4}^{2 -}$ is $96 u$
$∴$ Equivalent weight of $C u S O_{4} = 127 + 96 = 223 u$

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