[Saurabh]

150 mL of H2O2 sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M/2 H2SO4 for neutralization. Other part was treated with KMnO4 yielding 6.74 litre of O2 at STP. Using % yield indicated find volume strength of H2O2 sample used.       

H2O2 + 2KI ---------------> I2 + 2KOH
                 (40% yield)

H202 + 2KMnO4 + 3H2SO4 -----------> K2SO4 + 2MnSO4 + 3O2 + 4H20
                                      (50% yield)

[Babcock_Hall]

Sarah,

It is a forum rule (see the red link) that you must provide your thoughts before we can help you.