Question

To a 10 ml, 1M aqueous solution of $B{r}_2$ , excess of NaOH is added so that all $B{r}_2$ is disproportionated to $B{r}^{-}$ and $Br{O}_3^{-}$ . The resulting solution is made free of $B{r}^{-}$ by extraction and excess of $O{H}^{-}$ is neutralised by acidifying the solution. The resulting solution is sufficient to react with 2g of impure $Ca{C}_2{O}_4$ (M = 128 g/mole) sample. Which of the following is the correct % purity of oxalate sample

• A. 84.3%
• B. 32.5%
• C. 60%
• D. 64%

Answer 1

Answer: (D)

$6H^{+}+Br{O}_3^{-}+3C_2{O}_4^{2-}\to B{r}^{-}+6C{O}_2+3H_{2}O$

$10/3m$ mole $Br{O}_3^{-}$ reacts with $10m$. mole $C_2{O}_4^{2-}$

Mass of $Ca{C}_2{O}_4=10\times 10^{-3}\times 128=1.28g$

$\%$ purity $=\frac{\text{ Weight of substance }}{\text{ Weight of compound }}\times 100$ $\%$ purity $=\frac{1.28}{2}\times 100=64\%$