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 :

Options

A. 2.9
B. 29
C. 21
D. 4.9

Answer

A

Explanation

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`