Question

Calculate no. of moles and no. of atoms present in 5.6l of nh3

Answer 1

No. of moles (n) = (Volume of gas) ÷ (22.4 L of gas at STP)
So, n = 5.6/22.4
So, n = 1/4
So, n = 0.25 mole

1 mole of NH3 contains 6.022 × 10^23 molecules of NH3.

So, 0.25 mole of NH3 will contain
(6.022 × 10^23) molecules of NH3
i. e. 1.5055 × 10^23 molecules of NH3

Each molecule of NH3 contains 4 atoms
(1 atom of Nitrogen and 3 atoms of hydrogen)

So, No. of atoms (1.5055 × 10^23 molecules)
= 4 × 1.5055 × 10^23 atoms
= 6.022 × 10^23 atoms.