Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • 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

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Before NEET, CMC Vellore’s unique MBBS admissions tested aptitude along with merit; paper-leak restarts debate
anam-khan

NEET-UG paper leak fallout: Latur admin orders crackdown on illegal coaching classes
anam-khan

Re-NEET in 30 days a ‘massive logistical challenge’, NTA adopts ‘zero trust’ policy: Report

Latest Question