How many particles are present in 4.50g of caco3

0.271 × 10²³ particles

Explanation:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 4.50 g of CaCO₃:

100.1 g of CaCO₃ = 1 mole of CaCO₃ = 6.022 × 10²³ particles of CaCO₃

4.5 /100.1 = 0.045 mol

Particles in 0.045 mol:

0.045 ×6.022 × 10²³ particles

0.271 × 10²³ particles

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