50 ml of given H_(2)O_(2) solution is added to excess KI solution in acidic medium. The liberated I_(2) requires 20 ml of 0.04 M standard Hypo solution. Molarity of H_(2)O_(2) solution is

50 ml of given H_(2)O_(2) solution is added to excess KI solution in a

1.	50 ml of given H_(2)O_(2) solution is added to excess KI solution in acidic medium. The liberated I_(2) requires 20 ml of 0.04 M standard Hypo solution. Molarity of H_(2)O_(2) solution is
Answer» <html><body><p>`8xx10^(-3)M` <br/>`4xx10^(-3)M`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/5xx10-1901302" style="font-weight:bold;" target="_blank" title="Click to know more about 5XX10">5XX10</a>^(-3)M` <br/>None of these</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/2i-300397" style="font-weight:bold;" target="_blank" title="Click to know more about 2I">2I</a>^(-)+H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)+<a href="https://interviewquestions.tuteehub.com/tag/2h-300377" style="font-weight:bold;" target="_blank" title="Click to know more about 2H">2H</a>^(+)rarr2H_(2)O+I_(2)` <br/> `I_(2)+2S_(2)O_(3)^(-2)rarr2I^(-)+S_(4)O_(6)^(-2)` <br/> Eqts of `H_(2)O_(2)` = <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> `Na_(2)S_(2)O_(3)` <br/> `(2xx50)/(1000)xxM=(1xx20)/(1000)xx0.04impliesM=8xx10^(-3)`</body></html>