A solution of `H_(2)O_(2)` labelled as '20 V' was left open. Due to this some, `H_(2)O_(2)` decomposed and volume strength of the solution decreased . To determine the new volume strength of the `H_(2)O_(2)` solution, 10 mL of the solution was taken and it was diluted to 100 mL . 10 mL of this diluted solution was titrated against 25 mL of 0.0245 M `KMnO_(4)` solution under acidic condition. Calculate the volume strength of the `H_(2)O_(2)` solution .

Options

A. 15.00 V
B. 17.15 V
C. 20.00 V
D. 12.30 V

Answer

B

Explanation

Assuming new normality of original `H_(2)O_(2)` solution = X
After dilution to 100 mL of 10 mL of this solution, new normality will be (say `X_(1)`)
`:. X xx10=X_(1)xx100`
`X_(1)=(X)/(10) " " ...(i)`
10 mL of this dilute solution is titrated with 25 mL , 0.0245 M `KMnO_(4)` solution.
So, `N_(1)V_(1)=N_(2)V_(2)`
`(X)/(10)xx10=0.0245xx5xx25`
X=3.0625 N
So, volume strength of original `H_(2)O_(2)` solution `=X xx5.6 = 3.0625xx5.6 = 17.15`V