Question

Consider the following redox reaction:

$NaBr{O}_3+6H^{+}+6e^{-}\to NaBr+3H_{2}O$

Calculate the weight of sodium bromate $\left(NaBr{O}_3\right)$ required to prepare $0.6N$ of $100mL$ solution.
(Given : Molar mass of $NaBr{O}_3=151gmo{l}^{-1}$)

• A. $2g$
• B. $0.151g$
• C. $1.51g$
• D. $15.1g$

Answer 1

The correct option is C $1.51g$

Since,
$\text{Equivalents}=\text{Normality}\times \text{Volume (L)}$

$\text{Equivalents of}NaBr{O}_3=\frac{100\times 0.6}{1000}=0.06$

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

Change in oxidation state of $Br$ is from $+5$ in $NaBr{O}_3$ to $-1$ in $NaBr$.
$n_f=|+5-(-1)|\times 1n_f=6$

$\text{Equivalent weight of}NaBr{O}_3=\frac{M}{6}=\frac{151}{6}$
where M is the molar mass.


$\therefore \text{Equivalents of}NaBr{O}_3=\frac{W}{\frac{151}{6}}$

$\Rightarrow 0.06=\frac{W}{\frac{151}{6}}$


$\therefore W=1.51g$