Question

The weight of sodium bromate required to prepare $55.5 mL$ of $0.672 N$ solution for cell reaction,
$BrO _{3}^{-}+6 H ^{+}+6 e ^{-} \rightarrow Br ^{-}+3 H _{2} O ,$ is

• A. 1.56 g
• B. 0.9386 g
• C. 1.23 g
• D. 1.32 g

Answer 1

Answer: (B)

Meq. of $NaBrO _{3}=55.5 \times 0.672=37.296$

Let weight of $NaBrO _{3}= W$

$\therefore \frac{ W }{ M _{ NaBrO _{3}}} \times 6 \times 1000=37.296$

(equivalent weight $= M/6)$ of $n$-factor $= 6$

$\therefore \frac{ M }{151} \times 6 \times 1000=37.296$

$M = 0.9386\, g$