Question

How many total atoms are there in 29.1 g of water (H2O)?

Answer 1

As we know,

Molar mass of Hydrogen (H) = 1 g

Molar mass of Oxygen (O) = 16 g

Mass of Water present = 29.1 g    (Given)

So,

Molar Mass of Water (H₂O) = 2 * Molar mass of Hydrogen (H) + Molar mass of Oxygen (O)

Molar Mass of Water (H₂O) = 2 * 1 g + 16 g = 18 g

Now calculate no. of moles of water we are given,

No .of moles of Water (H₂O) given = Mass of Water (H₂O) given / Molar Mass of Water (H₂O)

No .of moles of Water (H₂O) given = 29.1 g / 18 g

No .of moles of Water (H₂O) given = 1.6167 mol

Now as we know,

No. of atoms in 1 mole of Water (H₂O) = 6.022 * 10²³ atoms   (Avogadro Number)

So,

No. of atoms in 1.6167 moles of Water (H₂O) = ( 1.6167 ) * ( 6.022 * 10²³ ) atoms

No. of atoms in 1.6167 mole of Water (H₂O) = 9.7355 * 10²³ atoms

Thus, in 29.1 g of Water (H₂O) 9.7355 * 10²³ atoms of water are present.