Question

The molarity of solution and weight of sodium bromate necessary to prepare $85.5$ mL of $0.672N$ solution when the half-cell reaction is $Br{O}_3^{⊝}+6H^{⨁}+6e^{-}\to B{r}^{⊝}+3H_{2}O$, are respectively :

• A. $M=0.112M$, Weight$=1.446mg$
• B. $M=0.224M$, Weight$=2.892g$
• C. $M=0.112M$, Weight$=1.446g$
• D. $M=1.112M$, Weight$=4.338g$

Answer 1

The correct option is C $M=0.112M$, Weight$=1.446g$

a) The reaction is shown below:
$Br{O}_3^{⊝}+6H^{⨁}+6e^{-}\Rightarrow B{r}^{⊝}+3H_{2}O$
The oxidation number of $Br$ changes from $+5$ to $-1$.
$n$ factor $=6$
Normality$=n\times$ molarity
Molarity$=\frac{0.672}{6}=0.112M$
Mass of $NaBr{O}_3=$ Molarity $\times V_{mL}\times {10}^{-3}$ x mol. mass $=0.112\times 85.5\times {10}^{-3}\times 151$ $=1.446g$
b) $2Br{O}_3^{⊝}+12H^{⨁}+10e^{-}\to B{r}_2+6H_{2}O$
The oxidation number of $Br$ changes from $+5$ to $0$.
$n$ factor $=5$
Normality $=n\times$ Molarity
Molarity$=\frac{0.672}{6}0.1344M$
Mass of $NaBr{O}_3$ = Molarity $\times V_{mL}\times {10}^{-3}$ x mol. mass $=0.1344\times 85.5\times {10}^{-3}\times 151$ $=1.735g$