Question

The weight of sodium bromate required to prepare 55.5 mL of 0.672 N solution,
$Br{O}_3^{-}+6H^{+}+6e^{-}\to B{r}^{-}+3H_{2}O$ is :

• A. 0.6186 g
• B. 0.9386 g
• C. 1.23 g
• D. 1.32 g

Answer 1

The correct option is B 0.9386 g

Equivalents of $NaBr{O}_3=\frac{55.5\times 0.672}{1000}=\frac{37.296}{1000}$

Let weight of $NaBr{O}_3=W$

As n factor is 6.

(Equivalent weight = M/6), where M is the molar mass.

M is the molar mass of $NaBr{O}_3=151g$

$\therefore \text{Equivalents of}NaBr{O}_3=\frac{W}{\frac{M_{NaBr{O}_3}}{6}}$

$\Rightarrow \frac{37.296}{1000}=\frac{W}{\frac{151}{6}}$

solving this,

$\therefore W=0.9386$ g

Hence, (b) is the correct answer.