Find the number of atoms present in 4.25g of NH3.

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Solution

Molecular mass of ammonia= 14+3=17

In 1 molecule (nh3) there are 1+3=4 atoms

1 mole (17 g) has Avogadro number of molecules

17 g=> 1 mole

4.25g=>4.25/17=0.25 moles

1 mole = 4×6.022×10^23 atoms

0.25 or 1/4 mole =6.022×10^23 atoms.

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Similar questions

The number of atoms present in 4.25 g of $({NH}_{3})$ is approximately.

a) $1 \times 10^{23}$

b) $1.5 \times 10^{23}$

c) $0.25 \times 10^{23}$

d) $6.02 \times 10^{23}$

4.25 g

{N H}_{3}

4.25 g

N H_{3}

4.25 g .

{N H}_{3}

N H_{3}

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