Question

Which of the foll contains greatest no of N atoms 1) 22.4 L nitrogen gas at STP 2) 500ml of 2.00 M of NO2 3) 1.00 mol of NH4Cl 4) 6.02 X 10 23 molecules of NO2

Answer 1

no of N gas:

1) in 22.4 l it is, equal to avogadro no. [6.022 * 10^23]

2)M=weight/(Atomic weight*volume)
weight=0.5*2=1gm 
for NO2, atomic weight=46, means the no. is (avogrado no.)/46

3) mol=weight/molar mass --> 6.022*10^23 
[why? because, weght=1.00*53=53gm] 

4)6.022*10^23

1, 3 and 4 seem to be the correct ones.