What weight of `FeSO_(4)` ( mol.wt. `=152`) will be oxidised by `200 mL` of normal `KMnO_(4)` solution in acid solution?

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

The volume of 10 volume H_(2)O_(2) solution that decolourises 200 mL of 2 N KMnO_(4) solution in acidic medium is

What is the equivalent weight of KMnO_(4) in acidic solution?

A 3 mole mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) required 100 mL of 2M KMnO_(4) solution in acidic medium. Find the mole of FeSO_(4) in the mixture.

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

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

10 ml of NaHC_2O_4 solution is neutralized by 10 ml of 0.1 M NaOH solution. 10 ml of same NaHC_2O_4 solution is oxidised by 10 ml of KMnO_4 solution in acidic medium. Calculate the molarity of KMnO_4 solution.