A sample of H_(2)O_(2) is x % by mass. If x ml of KMnO_(4) are required to oxidize 1 gm of this H_(2)O_(2) sample, calculate the normality of KMnO_(4) solution.

A sample of H_(2)O_(2) is x % by mass. If x ml of KMnO_(4) are require

1.	A sample of H_(2)O_(2) is x % by mass. If x ml of KMnO_(4) are required to oxidize 1 gm of this H_(2)O_(2) sample, calculate the normality of KMnO_(4) solution.
Answer» <html><body><p>`0.46N`<br/>`0.5N`<br/>`0.6N`<br/>`0.65N`</p>Solution :Let mass of `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)` solution = <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> gm <br/> mass of `H_(2)O_(2)` present is = <a href="https://interviewquestions.tuteehub.com/tag/xgm-3887063" style="font-weight:bold;" target="_blank" title="Click to know more about XGM">XGM</a> <br/> <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> that mass of `H_(2)O_(2)` solution taken = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> gm, <br/> So, mass of `H_(2)O_(2)` present in 1 gm solution <br/> `=(x)/(100)` Eq. of `H_(2)O_(2)=(x)/(100xx(34//2))` <br/> Eq. of `KMnO_(4)=NxxV(C)=Nxx x xx10^(-3)` <br/> Now `(x)/(100xx17)=Nxx x xx10^(-3)impliesN=0.59N`</body></html>