Question

In the given chemical reaction and find equivalent weight of caustic soda (in gram) ?
$I_2+NaOH\to NaI+NaI{O}_3+H_{2}O$

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

Answer 1

Answer: (C)

$3I_2+6NaOH\to \underset{n_{\text{factor }}=1}{5NaI}+\underset{n_{\text{factor }}=5}{1NaIO}$
$E_{NaOH}=\frac{M_{NaOH}}{n_{\text{factor }}}=\frac{40}{5/6}=48g$