Observe the following reaction: I_2 + 10 HNO_ | vedclass

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(A) In the given reaction: $I_2 + 10 HNO_3 \rightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$. First,determine the change in the oxidation state of nitrogen in $

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$M$

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(A) In the given reaction: $I_2 + 10 HNO_3 ightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$. First,determine the change in the oxidation state of nitrogen in $HNO_3$. In $HNO_3$,the oxidation state of $N$ is $+5$. In $NO_2$,the oxidation state of $N$ is $+4$. The change in oxidation state per molecule of $HNO_3$ is $|5 - 4| = 1$. Since the change in oxidation state (n-factor) for $HNO_3$ is $1$,the equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}} = \frac{M}{1} = M$.

Observe the following reaction: $I_2 + 10 HNO_3 \rightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$. The equivalent weight of $HNO_3$ is (molar mass of $HNO_3 = M$).

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