Observe the following reaction $\mathrm{I}_2+10 \mathrm{HNO}_3 \rightarrow 2 \mathrm{HIO}_3+10…
Observe the following reaction
$\mathrm{I}_2+10 \mathrm{HNO}_3 \rightarrow 2 \mathrm{HIO}_3+10 \mathrm{NO}_2+4 \mathrm{H}_2 \mathrm{O}$
The equivalent wt. of $\mathrm{HNO}_3$ is $\left(\right.$ molar mass of $\left.\mathrm{HNO}_3=\mathrm{M}\right)$
M
$\frac{M}{4}$
$\frac{M}{2}$
$\frac{M}{5}$
Solution
$\mathrm{I}_2+10 \mathrm{HNO}_3 \rightarrow 2 \mathrm{HIO}_3+10 \mathrm{NO}_2+4 \mathrm{H}_2 \mathrm{O}$
Equivalent wt. $=\frac{\text { Molar mass }}{\text { no. of } \mathrm{H}^{+} \text {produced by acid }}=\frac{\mathrm{M}}{1}=\mathrm{M}$