Answer : C - 1.02 x 10²³
Explanation ;
1 molecule of Al(NO₃)₃ has 3 N atoms.
Al(NO₃)₃ → Al + 3N + 9O
Moles (mol) = Mass (g) / Molar mass (g/mol)
Mass of Al(NO₃)₃ = 12 g
Molar mass of Al(NO₃)₃ = (27 + 14 x 3 + 16 x 9) g/mol
= 213 g/mol
Mass of Al(NO₃)₃ = 12 g / 213 g/mol
= 0.0563 mol
The stoichiometric ratio between Al(NO₃)₃ compound and N atoms is 1 : 3.
Hence,
the moles of N atoms = moles of Al(NO₃)₃ x 3
= 0.0563 mol x 3
= 0.1689 mol
According to the Avogadro constant,
1 mol of substance = 6.02 x 10²³ particles
Hence, N atoms in the 12 g of Al(NO₃)₃ = 0.1689 mol x 6.02 x 10²³ mol⁻¹
= 1.016778 x 10²³
= 1.02 x 10²³
