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

• A

  84.3%
• B

  32.5%
• C

  60%
• Correct answer
  D

  64%

Solution

$6{\mathrm{H}}^{+}+\mathrm{B}\mathrm{r}{\mathrm{O}}_3^{-}+3{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to \mathrm{B}{\mathrm{r}}^{-}+6\mathrm{C}{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$
10/3 m mole $\mathrm{B}\mathrm{r}{\mathrm{O}}_3^{-}$ reacts with 10 m. mole ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$
Mass of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\text{10}\text{×}{\text{10}}^{-\text{3}}\text{×}\text{128}\text{=}\text{1}\text{.28}\text{g}$
$\text{\%}\text{purity}\text{=}\frac{\text{Weight}\text{of}\text{substance}}{\text{Weight}\text{of}\text{compound}}\text{×}\text{100}$
$$ \%\text{purity}=\frac{1.28}{2}\times 100=64\text{\%}$$

Hence, the correct answer is option: 'D'.