A 36.04 g sample of water is boiled. When the steam expands to fill the vessel how many atoms of hydrogen would be found in the sample?

The number of hydrogen atoms in 36.04 g  sample of steam is 2.408 × 10²⁴ atoms of hydrogen

Explanation:

The  given parameters are;

The mass of the sample of water, H₂O = 36.04 g

From the principle of conservation of matter, the mass of the water = The mass of the steam, therefore;

The mass of the steam = 36.04 g

The molar mass of H₂O = 18.01528 g/mol

Therefore, the number of moles, n, of H₂O in the 36.04 g of steam is given as follows;

n = Mass/(Molar Mass) = (36.04 g)/(18.01528 g/mol) = 2.00052399963 ≈ 2 moles

The number of molecules of H₂O per mole of H₂O is geven by the Avogadro's number = 6.02 × 10²³ molecules

Therefore;

The number of H₂O in 2 moles of H₂O = 2 × 6.02 × 10²³ = 1.204 × 10²⁴ molecules

The number of hydrogen atoms per molecule of H₂O = 2 hydrogen atoms

Therefore;

The number of hydrogen atoms in the 1.204 × 10²⁴ molecules of H₂O = 2 × 1.204 × 10²⁴ molecules of H₂O

Which gives;

The number of hydrogen atoms in the 1.204 × 10²⁴ molecules of H₂O = 2.408 × 10²⁴ atoms of hydrogen

Therefore;

The number of hydrogen atoms in 36.04 g  sample of steam = 2.408 × 10²⁴ atoms of hydrogen.