Question

In the given chemical reaction and find equivalent weight of caustic soda (in gram) ?
$I_{2}+NaOH \rightarrow NaI+NaIO_{3}+H_{2}O$

• A. $40$
• B. $\frac{1}{40}$
• C. $48$
• D. $\frac{40}{6}$

Answer 1

Answer: (C)

$3I_{2}+6NaOH \rightarrow \underset{n_{\text{factor }} = 1}{5 NaI}+\underset{n_{\text{factor }} = 5}{1 NaIO}$
$E_{NaOH}=\frac{M_{NaOH}}{n_{\text{factor }}}=\frac{40}{5 / 6}=48g$