Question
A 0.5 g sample of sodium nitrate on heating gives 44.8ml of O2 at STP. % purity of the sample is.............
my answer is 64% but the correct answer is 68%. HOW???
Joshi sir comment
NaNO3 −−−−> NaNO2 + 1/2 O2
molar mass of sodium nitrate = 85
so 85 gm will give 11200 ml O2
so x gm will give 11200*x/85 = 2240x/17
on comparing with 44.8
we get 2240x/17 = 44.8
so x = 44.8*17/2240 = 17/50 = 34/100 = 0.34 gm
means 0.34 gm sodium nitrate is pure in 0.5 gm. sample so % purity = 0.34*100/0.50 = 68%