Question

calculate the number of atom of oxygen present in 120g of nitric acid

Answer 1

Compound HNO³:

G.M.M = 1+ 14+ 3*16= 63

63 g of HNO³ contains= 6.02* 10²³ atoms

120g of HNO³ contains
=(120/ 63) * 6.02* 10²³ atoms

No. of oxygen atoms
= 3* (120/63) * 6.02*10²³ atoms = 3.44 * 10²⁴ atoms.

Hope it helps ^_^

Answer 2

Molar mass of HNO3 = 63g

63g of HNO3 = 1 mole of HNO3

1 g of HNO3 = 1/63 moles of HNO3

So,
      120g of HNO3 = 120/63 moles of HNO3 
                              = 1.905 moles of HNO3 

1 mole of the substance = 6.022 x 10²³ molecules of the substance 

So,
     1 mole of HNO3 = 6.022 X 10²³ molecules of HNO3

Therefore, 
               1.9065 moles of HNO3 = 1.905 x 6.022 x 10²³ molecules of HNO3 
                                                    = 1.147 x 10^24 molecules of HNO3

1 molecules of HNO3 contains three atoms of oxygen. 

So,
         1.147 x 10^24 molecules of HNO3 will contains 
                      = 3 x 1.147 x 10^24 atoms of oxygen
                      = 3.442 x 10^24 atoms of oxygen.

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Hope it helps ^_^