question_answer

\[{{H}_{2}}{{O}_{2}}+2KI\xrightarrow{40\,(percent)\,yield}\,{{I}_{2}}\,+2KOH\] \[{{H}_{2}}{{O}_{2}}+2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{50\,(percent)\,yield}\,{{K}_{2}}S{{O}_{4}}+\] \[2MnS{{O}_{4}}+3{{O}_{2}}+4{{H}_{2}}O\] 150 mL of \[{{H}_{2}}{{O}_{2}}\] sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of \[M/2\,\,{{H}_{2}}S{{O}_{4}}\] for neutralisation. Other part was treated with \[KMn{{O}_{4}}\] 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.

Answer:

33.6