`H_2O_2+2KI rarr 40% yield I_2 + 2KOH` `H_2O_2+2KMnO_4+3H_2SO_4 rarr 50% yield` `K_2SO_4+2KMnSO_4+3O_2 +4H_2O` `150mL` of `H_2 O_2` sample was divided into two parts. First part was treated with `KI` and formed `KOH` required `200mL` of `M/2H_2SO_4` for neutralisation. Other part was treated with `KMnO_4` yielding `6.74` litre of `O_2` at 1 atm. and `273K`. Using % yield indicated find volume strength of `H_2O_2` sample used.

H_(2)O_(2)+2KIoverset(40% "yield")rarrI_(2)+2KOH H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)overset(50% "yield")rarr K_(2)SO_(4)+2MnSO_(4)+3O_(2)+4H_(2)O 150mL of H_(2)O_(2) sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M//2H_(2)SO_(4) for neutralisation. Other part was treated with KMnO_(4) yielding 6.74 litre of O_(2) at 1 atm. and 273 K. Using % yield indicated find volume strength of H_(2)O_(2) sample used.

200 ml of 0.01 M KMnO_(4) oxidise 20 ml of H_(2)O_(2) sample in acidic medium. The volume strength of H_(2)O_(2) is

0.2 g if a sample of H_2 O_2 required 10 ml of 1 N KMnO_4 for titration in acidic medium. The percentage purity of H_2 O_2 sample is:

80 mL of M/24 K_(2)Cr_(2)O_(7) oxidises 22.4 mL H_(2)O_(2) solution. Find volume strength of H_(2)O_(2) solution.

A : H_2O liberate O_2 an reaction with KMnO_4 in acidic medium. R : KMnO_4 oxidises H_2O_2" to " O_2 .

10 g sample of H_(2)O_(2) just decolorised 100 ml of 0.1 M KMnO_(4) in acidic medium % by mass of H_(2)O_(2) in the sample is

Balance the equations given below : K_2Cr_2O_7 + H_2SO_4 +SO_2 to K_2SO_4 + Cr_2(SO_4)_3 + H_2O

Na_(2)SO_(4).xH_(2)O has 50% H_(2)O . Henxe, x is :

The purity of H_2O_2 in a given sample is 85%. Calculate the weight of impure sample of H_2O_2 which requires 10mL of M//5 KMnO_4 solution in a titration in acidic medium

The purity of H_2O_2 in a given sample is 85%. Calculate the weight of impure sample of H_2O_2 which requires 10mL of M//5 KMnO_4 solution in a titration in acidic medium