AlCl3 = Al3+ + 3Cl- = 4 ions
1 mol AlCl3 = Al3+ + 3Cl- = 4 x 6.023 x 10^23 ions
Hence, 1.5 moles of Aluminum Chloride, AlCl3 = 1.5 x 4 x 6.023 x 10^23 ions = 36.14 x 1023 ions
Therefore, 36.14 x 10^23 ions are there in 1.5 moles of Aluminum Chloride, AlCl3
