Al2(SO4)3 solution of 1molal concentration is present in 1 liter solution of 2.684 g/cc. How many moles of BaSO4 would be precipitated on adding BaCl2 in excess ?

Let no. of moles of Al2(SO4)3 in solution is xnsolutewt. of solvent×1000=1x2684−342x×1000=1 x= 2Al2(SO4)3+3BaCl2→3BaSO4+2AlCl32Al2SO43⟶6 moles SO42−∴6 moles of BaSO4