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  • For disproportionation of 20 mL, 1 M  solution, excess of <img alt="\mathrm{Na

For disproportionation of 20 mL, 1 M \mathrm{Br_{2}} solution, excess of \mathrm{NaOH} used. By extraction, the solution so formed do not contain \mathrm{Br}^{-}and by addition of acid the exes of \mathrm{OH}^{-}was neutralised. The final solution is enough to react with 3 g of impure \mathrm{\mathrm{CaC}_2 \mathrm{O}_4(\mathrm{M}=128 \mathrm{~g} / \mathrm{mol})} sample. Find % purity of sample of oxalate.

Option: 1

85.3 %


Option: 2

12.5 %


Option: 3

90 %


Option: 4

84 %


Answers (1)

best_answer

3 \mathrm{Br}_2+6 \mathrm{OH}^{-} \rightarrow 5 \mathrm{Br}^{-}+\mathrm{BrO}_3^{-}+3 \mathrm{H}_2 \mathrm{O}\text { ...........(1) }

6 \mathrm{H}^{+}+\mathrm{BrO}_3^{-}+3 \mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow \mathrm{Br}^{-}+6 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \text { .......(2) }

20 m mol of \mathrm{Br_{2}} produces \frac{20}{3}\text{m} mol \mathrm{BrO_{3}^{-}}

Required m moles of \mathrm{CaC}_2 \mathrm{O}_4=\frac{20}{3} \times 3=20

\therefore mass of \mathrm{CaC}_2 \mathrm{O}_4=20 \times 10^{-3} \times 128

% purity =\frac{20 \times 10^{-3} \times 128}{3} \times 100

               = 85.3 %

Posted by

Devendra Khairwa

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