25 ml of H_(2)O_(2) solution were added to excess of acidified KI solution. The iodine so liberated required 20 ml of 0.1 N Na_(2)S_(2)O_(3) solution. Calculate strength in terms of normality and percentage.

25 ml of H_(2)O_(2) solution were added to excess of acidified KI solu

1.	25 ml of H_(2)O_(2) solution were added to excess of acidified KI solution. The iodine so liberated required 20 ml of 0.1 N Na_(2)S_(2)O_(3) solution. Calculate strength in terms of normality and percentage.
Answer» <html><body><p>0.04 N, `0.136%`<br/>0.08 N, `0.136%`<br/>0.08 N, `0.163%`<br/>0.02 N, `0.163%`</p>Solution :`O_(2)^(-)+<a href="https://interviewquestions.tuteehub.com/tag/2e-300683" style="font-weight:bold;" target="_blank" title="Click to know more about 2E">2E</a>^(-) rarr 2O^(-2),"" 2I^(-) rarr I_(2)+2e^(-)` <br/> `2(S^(+2))_(2) rarr (S^(+5//2))_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)+2e^(-), I_(2)+2e^(-) rarr 2I^(-)` <br/> Meq. Of `H_(2)O_(2)=` Meq. of `I_(2)=` Meq. of `Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `w/(34//2) times 1000=20 times 0.1` <br/> `therefore W_(H_(2)O_(2))=0.034 gr//25ml` <br/> `therefore N_(H_(2)O_(2))=0.034/(34//2) times 1000/25=0.08` <br/> Vol. <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a> `=5.6 times 0.08=0.448` <br/> % strength `=17/56 times 5.6 times 0.08=0.136%`</body></html>