What is the mass of hydrogen peroxide present in 1 litre of a 2M solution ? Calculate the volume of oxygen at S.T.P. liberated upon the complete decomposition at 100 `cm^3` of the above solution.

Explanation

Step 1. To calculate the mass of `H_(2)O_(2)` in 1 litre of 2M solution.
Molecular mass of `H_(2)O_(2)=2xx1+2xx16=34` u.
By definition, 1 litre of 1 M `H_(2)O_(2)` contains 34 g of `H_(2)O_(2)`
`therefore` 1 litres of 2M `H_(2)O_(2)` will contain `H_(2)O_(2)=34xx2=68`
Step 2. To calculate the volume of `O_(2)` liberated at STP from `100 cm^(3)` of 2M solution.
1 litre of 2M `H_(2)O_(2)` solution will contain `H_(2)O_(2)=(68)/(1000)xx100=6.8g`
The equation representing the decomposition of `H_(2)O_(2) ` is
`underset(2xx34=68 g)(2H_(2)O_(2))to 2H_(2)O + underset(22400 cm^(3) at "STP")(O_(2))`
Now 68 g of `H_(2)O_(2)` at STP give `O_(2)=22400 cm^(3)`
`therefore 6.8 ` g of `H_(2)O_(2)` at STP will evolve `O_(2)=(22400)/(68)xx6.8=2240 cm^(3)=2.24 litres.`