H_2O_2+ 2KI overset("40% yield")to I_2 + 2KOHH_2O_2 + 2KMnO_4 + 3H_2SO_4 overset("50% yield ")to K_2SO_4 + 2MnSO_4 + 3O_2 + 4H_2O 150 ml of H_2O_2sample was divided into two parts. First part was treated with KI and Formed KOH required 200 ml. of M//2 H_2SO_4 for neutralisation.Other part was trated with KMnO_4 yielding 6.74 litre of O_2 at STP.Using % yieldindicated find volume stregth of H_2O_2 sample used.

H_2O_2+ 2KI overset("40% yield")to I_2 + 2KOHH_2O_2 + 2KMnO_4 + 3H_2SO

1.	H_2O_2+ 2KI overset("40% yield")to I_2 + 2KOHH_2O_2 + 2KMnO_4 + 3H_2SO_4 overset("50% yield ")to K_2SO_4 + 2MnSO_4 + 3O_2 + 4H_2O 150 ml of H_2O_2sample was divided into two parts. First part was treated with KI and Formed KOH required 200 ml. of M//2 H_2SO_4 for neutralisation.Other part was trated with KMnO_4 yielding 6.74 litre of O_2 at STP.Using % yieldindicated find volume stregth of H_2O_2 sample used.
Answer» 5.04 1.08 3.36 11.33 Solution :No. of GEW .s of `H_2SO_4` = No.of GEW.s of KOH = No. of GEW.s `H_2O_2` = 0.2 No. of GEW.s `H_2O_2` = No. of GEW.s of `O_2 = 6.74/22.4 XX 2 = 0.6` in part I 0.4 X = 0.2 X = 0.5 in part II 0.5 X = 0.6 X =1.2 (where X is No. of GEW.s `H_2O_2`) TOTAL No. of GEW.s of `H_2O_2` `=1.7 N = 1.7 xx 100/15` volume stregth = 11.33