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Calculate the number of aluminum ions in `250 cm^3` of 0.1 M aluminium sulphate (Avogadro’s Constant = `6.0 times 10^23`)

Answer Text:
Moles `Al_2(SO_4)_3 = 0.1 times 250/1000 = 0.025`
Moles `Al^(3+) = 3 times 0.025 = 0.075`
No. `Al^(3+)` ions =`0.075 times 6.0 times 1023 = 4.5 times 10^22`