In the reaction 2CuSO_4 + 4KI rightarrow Cu_2 | vedclass

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(D) In the given reaction: $2CuSO_4 + 4KI \rightarrow Cu_2I_2 + 2K_2SO_4 + I_2$. The oxidation state of $Cu$ in $CuSO_4$ is $+2$. The oxidation state

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$1$

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(D) In the given reaction: $2CuSO_4 + 4KI ightarrow Cu_2I_2 + 2K_2SO_4 + I_2$. The oxidation state of $Cu$ in $CuSO_4$ is $+2$. The oxidation state of $Cu$ in $Cu_2I_2$ (which is $CuI$) is $+1$. The change in oxidation state per $Cu$ atom is $2 - 1 = 1$. Since there is one $Cu$ atom in $CuSO_4$,the $n$-factor for $CuSO_4$ is $1$. The equivalent weight is given by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{n	ext{-factor}}$. Therefore,$\frac{	ext{Equivalent weight}}{	ext{Molecular weight}} = \frac{1}{n	ext{-factor}} = \frac{1}{1} = 1$.

In the reaction $2CuSO_4 + 4KI \rightarrow Cu_2I_2 + 2K_2SO_4 + I_2$,what is the ratio of the equivalent weight of $CuSO_4$ to its molecular weight?

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