Question

What is the mass of Al2(SO4)3 needed to prepare 40 (p3)ml of (0.2p2)M of sulphate ions in aluminum sulphate solution? where p= log(moles of Al2(SO4)3 ?

Answer 1

$Al_2(SO_4)_3$ =342.15 g/mol

$Moles =volume×Molarity$

$Moles =0.04×0.2=0.008moles$

$Mass =0.008×342.15=2.74grams$