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50 mL of an aqueous solution of H_(2)O_(2) was reacted with excess of KI solution and dilute H_(2)SO_(4) The liberated iodine requried 20 mL 0.1 N Na_(2)S_(2)O_(3) solution for complete interaction calculate the concentration of H_(2)O_(2) in gL^(-1)

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</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)^(-1)KI+H_(2)SO_(4)rarrK_(2)SO_(4)+I_(2)+2H_(2)O^(-2)` <br/> `Na_(2)S_(2)O_(3)+I_(2)rarrNa_(2)S_(4)O_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)+2` Nal <br/> Total change in O.N of O=2(-1)-2(-2)=2 `therefore` Eq of `H_(2)O_(2)=(2+2xx16)/(2)=<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>` <br/> Let `N_(1)` be the normality of `I_(2)` solution <br/> Since one equivalent of `H_(2)O` produces 1 equilvent of `I_(2)` <br/> `therefore` 50mL of `N_(1) I_(2)` solution =50mL of `N_(1)H_(2)O_(2)` solution <br/> `therefore N_(1)=(20xx0.1)/(50)=0.04` N or <a href="https://interviewquestions.tuteehub.com/tag/streangth-3079918" style="font-weight:bold;" target="_blank" title="Click to know more about STREANGTH">STREANGTH</a> of `H_(2)O_(2) "solution" =0.04xx17=0.68 gL^(-1)`</body></html>


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