Question

$18$ mL of $1.0$M $B{r}_2$ solution undergoes complete disproportionation in basic medium to $B{r}^{⊝}$ and $Br{O}_3^{⊝}$. Then, the resulting solution requires $45$ mL of $A{s}^{3+}$ solution to reduce $Br{O}_3^{⊝}$ to $B{r}^{⊝}$. $A{s}^{3+}$ is oxidized to $A{s}^{5+}$. Which statements are correct?

• A. Equivalent weight of $B{r}_2=\frac{M}{10}$
• B. Equivalent weight of $B{r}_2=\frac{5M}{3}$
• C. Molarity of $A{s}^{3+}=0.4M$
• D. Molarity of $A{s}^{3+}=0.2M$

Answer 1

Answer: (B), (C)

Equivalent weight of $B{r}_2=\frac{5M}{3}$
Molarity of $A{s}^{3+}=0.4M$

The balanced chemical equations are $3B{r}_2+6O{H}^{-}\to Br{O}_3^{-}+5B{r}^{-}+3H_{2}O$
$3A{s}^{3+}+Br{O}_3+3H_{2}O\to 3A{s}^{5+}+B{r}^{-}+6O{H}^{-}$
In the first equation the oxidation number of bromine changes from $0$ in $B{r}_2$ to +5 in $Br{O}_3^{-}$.
Thus, the change in the oxidation number for 3 bromine molecules is $+5$.
Change in the oxidation number for one bromine molecule is $\frac{+5}{3}$.
Hence, the equivalent weight of bromine is $\frac{5M}{3}$, where, M is the molecular weight of bromine.
The number of moles of bromine is $\frac{18}{1000}\times 1.0=0.018mol$.
From the balanced chemical equations,
$3molB{r}_2=1molBr{O}_3^{-}=3molA{s}^{3+}$
Hence, the number of $A{s}^{3+}=0.018mol$.
Hence, the molrity of $A{s}^{3+}=\frac{0.018\times 1000}{45}=0.40mol$.