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