10 grams of a sample of potassium chlorate gave on complete decomposition 2.24 litre of oxygen at N.T.P. What is the percentage purity of the sample ?

Answer

`81.7 %`

Explanation

Chemical equation for the reaction is :
`underset(underset(=245g)(2xx(39+35.5+48)))(2KClO_(3)(s))overset("Heat")(rarr)2KCl(s)+underset(underset(=67.2L)(3xx22.4))(3O_(2)(g))`
67.2 L of `O_(2)` is evolved at N.T.P from `KClO_(3) 245 g`
2.24 L of `O_(2)` is evolved at N.T.P from `KClO_(3)=(245)/(67.5)xx2.24=8.167g`
Percentage purity of sample `= ("Mass of pure "KClO_(3))/("Mass of the sample")xx100=(8.167)/(10.0)xx100=81.7%`.