The number of atoms present in 4.25 g of NH3 is approximately 6.02 x 10^23. (a). 1 x 10^23 (b). 1.5 x 10^23

By Shivank Goel|Updated : August 10th, 2022

It is given that

Mass of ammonia = 4.25 g

Gram molecular mass of ammonia = 17 g

We know that

Number of moles of ammonia = Mass of ammonia/ Gram molecular mass of ammonia

Substituting the values

= 4.25/17

= 0.25 mol

1 mole of ammonia has 6.022 x 1023 molecules of ammonia

So the number of molecules present in 0.25 mol of ammonia = 0.25 x 6.022 x 1023

= 1.505 x 1023 molecules

There are 4 atoms in 1 molecule of ammonia (3 hydrogens and 1 nitrogen)

Number of atoms present in 1.505 x 1023 ammonia molecules = 4 x 1.505 x 1023

= 6.02 x 1023

Therefore, the number of atoms present is 1 x 1023.

Summary:

The number of atoms present in 4.25 g of NH3 is approximately 6.02 x 1023.

  1. 1 x 1023

  2. 1.5 x 1023

The number of atoms present in 4.25 g of NH3 is approximately 6.02 x 1023 is 1 x 1023.

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