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

Explanation

(i) Molar mass of `H_(2)O_(2) = 34 g mol^(-1)`
1 L of 5 M solutions of `H_(2)O_(2)` will contain `34 xx 5 g H_(2)O_(2)`
2 L of 5 M solutions of `H_(2)O_(2)` will contain `34 xx 5 xx 2 = 340 g H_(2)O_(2)`
Mass of `H_(2)O_(2)` present in 2L of 5 mole solution = 340 g
(ii) 0.2 L (ou 200 mL) of 5 M solutions will contain
`underset(2)overset(340xx0.2)to = 34 g H_(2)O_(2)`
`underset(2xx34xx68g)(2H_(2)O_(2)) to 2H_(2)O + underset(2xx16=32g)(O_(2))`
`68 g H_(2)O_(2)` on decomposition will give 32 g `O_(2)`
`therefore 34 g H_(2)O_(2)` on decomposition will give `(32xx34)/(68) = 16 g O_(2)`