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» 2.9 29 21 4.9 Solution :Number of moles of `KMnO_(4)=(MV)/(1000)=(0.145xx46.9)/(1000)` `=6.8xx10^(-3)` Number of moles of `H_(2)O_(2)=6.8xx10^(-3)xx2.5=0.017` Mass of `H_(2)O_(2)=0.017xx34=0.578` Mass % of `H_(2)O_(2)=(0.578)/(20)xx100=2.9`