100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume of Oxygen in (MnO_4^(-)) changes to Mn^(2+)in acidic medium and to MnO_4 in alkaline medium).

100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume

1.	100 mL of 0.01M KMnO_4 oxidises 100mL of acidified H_2O_2 . The volume of Oxygen in (MnO_4^(-)) changes to Mn^(2+)in acidic medium and to MnO_4 in alkaline medium).
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/26-298265" style="font-weight:bold;" target="_blank" title="Click to know more about 26">26</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a><br/>260 mL<br/>`2.6 mL`<br/>26 Lt</p>Solution :`(N_(1)V_(1))_(KMnO_(4))=(N_(2)V_(2))_(H_(2)O_(2))` <br/> `N=0.05 rArr 11.2` <a href="https://interviewquestions.tuteehub.com/tag/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a> of `H_(2)O_(2)=(11.2xx0.05)/(2)` <br/> Vol of `O_(2)=(11.2xx0.05)/(2)xx100 mL`</body></html>