What is the approximate volume (in mL) of 10 vol `H_2O_2` solution that will react completely with 1 L of 0.02 M `KMnO_4` solution in acid medium ?

Options

A. 56.05
B. 113.5
C. 90.8
D. 75.75

Answer

A

Explanation

Given ,
10 vol of `H_2O_2` solution
i.e., 1 L of `H_2O_2` contains 10 L of `O_2` at STP
Volume of `KMnO_4` = 1L
Molarity of `H_2O_2` = 0.02 M
The reaction between `KMnO_4` and `H_2O_2` is given by ,
`2KMnO_4 + 5H_2O_2 + 3H_2SO_4 to 5O_2 + 2MnSO_4 + K_2SO_4 + 8 H_2O`
`rArr` 2 moles of `KMnO_4` = 5 moles of `H_2O_2` = 5 moles of `O_2`
Thus , at STP,
34g of `H_2O_2`= 11.2 L of `O_2`
Thus , 34 x 5 g of `H_2O_2 = (34xx5xx11.2)/34`
`rArr` 1mL of `H_2O_2` = 56ml of `O_2`
`therefore` Volume of `H_2O_2` = 56 ml