A solution of H_(2)O_(2) is titrated against a solution of KMnO_(4). The reaction is : 2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) to 2Mn^(2+)+5O_(2)+8H_(2)O If it requires 46.9 mL of 0.145 M KMnO_(4) to oxidise 20 g of H_(2)O_(2), the mass percentage of H_(2)O_(2) in this solution is :

A solution of H_(2)O_(2) is titrated against a solution of KMnO_(4).

1.	A solution of H_(2)O_(2) is titrated against a solution of KMnO_(4). The reaction is : 2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) to 2Mn^(2+)+5O_(2)+8H_(2)O If it requires 46.9 mL of 0.145 M KMnO_(4) to oxidise 20 g of H_(2)O_(2), the mass percentage of H_(2)O_(2) in this solution is :
Answer» <html><body><p>2.9<br/>29<br/>21<br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>.9</p>Solution :Number of moles of `KMnO_(4)=(MV)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=(0.145xx46.9)/(1000)` <br/> `=6.8xx10^(-3)` <br/> Number of moles of `H_(2)O_(2)=6.8xx10^(-3)xx2.5=0.017` <br/> Mass of `H_(2)O_(2)=0.017xx34=0.578` <br/> Mass % of `H_(2)O_(2)=(0.578)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)xx100=2.9`</body></html>