How many ions are present in 9.4g of pottasium oxide K2O? (K=39, O=16) (Avogadros constant=6.0*10^23/mol)

1) k2O = 2K (1+) + O (2-)

This is, every molecule of K2O consists of three ions (2 of K 1+, and one of O 2-)

2) Caluclate the number of moles of K2O

- molar mass of K2O = 2*39 + 16 = 94 g/mol

- n = mass / molar-mass = 9.4g / 94 g/mol = 0.1 mol

3) Multiply by 3 to have the number of moles of ions: 3 * 0.1 mol = 0.3 mol of ions.

4) Multiply by Avogadro's number to obtain the number of ions"

0.3 mol * 6.0*10^23 = 1.8 * 10^23 ions.

Answer: 1.8*10^23 ions