A solution containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions is titrated with 20 " mL of " `(M)/(4)` `KMnO_(4)` solution in acidic medium. The resulting solution is treated with excess of KI after neutralisation. The evolved `I_(2)` is then absorbed is 25 " mL of " `(M)/(10)` hypo solution. Which of the following statements are correct?

25 mL of 0.50M H_(2)O_(2) solution is added to 50 mL of 0.20 M KMnO_(4) is acid solution. Which of the following statements is true?

3 " mol of "a mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) requried 100 " mL of " 2 M KMnO_(4) solution in acidic medium. Hence, mole fraction of FeSO_(4) in the mixture is

20 ml of 0.1 M FeC_(2)O_(4) solution is titrated with 0.1 M KMnO_(4) is acidic medium. Volume of KMnO_(4) solution required to oxidise FeC_(2)O_(4) completely is

Calculate the volume of 10 volume H_(2)O_(2) solution that will react with 200 mL of 2N KMnO_(4) in acidic medium.

10 mL of 1 NHCl is mixed with 20 mL of 1 MH_(2)SO_(4) and 30 mL of 1M NaOH. The resultant solution has:

If 1.34g Na_(2)C_(2)O_(4) is dissolved in 500 mL of water and this solution is titrated with M/10 KMnO_(4) solution in acidic medium, the volume of KMnO_(4) used is :

250 ml of 0.10 M K_2 SO_4 solution is mixed with 250 ml of 0.20 M KCI solution. The concentration of K^(+) ions in the resulting solution will be:

A 0.46 g sample of As_(2)O_(3) required 25.0 " mL of " KMnO_(4) solution for its titration. The molarity of KMnO_(4) solution is

100 mL of NaHC_(2)O_(4) required 50 mL of 0.1M KMnO_(4) solution in acidic medium. Volume of 0.1M NaOH required by 100 mL of NaHC_(2)O_(4) is :

A solution of 0.2 g of a compound containing Cu^(2+) and C_(2)O_(4)^(2-) ions on titration with 0.02M KMnO_(4) in presence of H_(2)SO_(4) consumes 22.6mL oxidant. The resulting solution is neutralized by Na_(2)CO_(3) , acidified with dilute CH_(3)COOH and titrated with excess of KI . The liberated I_(2) required 11.3 mL "of" 0.05M Na_(2)S_(2)O_(3) for complete reduction. Find out mole ratio of Cu^(2+) and C_(2)O_(4)^(2+) in compound.