How many atoms of each element are present in 6.3g of nitric acid (HNO3) ?

Molecular Mass of HNO3 = (1 + 14 + 16X3)gm

= 63 gm

Therefore 63 gm HNO3=1 mole
1 g of nitric acid = 1/63 moles
6.3 mole = 0.1 mole
As 1 mole = 6.022 X 1023 particles 

 0.1 mole = 6.022 X 10 23 X 0.1  molecule
Hence 6.3 g of nitric acid contains 6.022 X 10 22 molecules 

One molecule of HNO3 contain 1 hydrogen, 1 nitrogen and 3 oxygen atoms
Therefore, 
In 6.3 g of nitric acid, 6.022 X 10 22 atoms of hydrogen, 6.022 X 10 22 atoms of nitrogen and 18.06  X 10 22 atoms of oxygen are present. 

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