(i) What mass of hydrogen peroixde will be present in 2 L of a 5 molar solution ?
(ii) Caluclate the mass of oygen which will be liberated by the decomposition of 200 mL of this solution.

Explanation

Mol. Mass of `H_(2)O_(2)=34 g "mol"^(-1)`
`therefore 1 L` of M solution of `H_(2)O_(2)` will contain `H_(2)O_(2)=34xx5` g
`therefore 2 `L of 5 M solution will contain `H_(2)O_(2)=34xx5xx2=340 g`
or 200 mL of 5 solution will contain `H_(2)O_(2)=(340)/(2000)xx200=34` g
Now `underset(68 g)(2H_(2)O_(2))to 2H_(2)O + underset(32g)(O_(2))`
Now 68 g of `H_(2)O_(2)` on decomposition will give `O_(2)=32 g`
`therefore 34 g ` of `H_(2)O_(2)` on decomposition will give `O_(2)=(32)/(68)xx34=16` g