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

Dear Student,

The density of the solution is 2.684 g/cc, which includes both the solute Al₂(SO₄)₃ and the solvent water. So, deducing the density of water which is ~ 1 g/cc, we will have 1.684 g/cc of ​Al₂(SO₄)₃ .

So, in a litre of solution we have 1.684 x 1000 = 1684 g of Al₂(SO₄)₃ .

Moles of Al₂(SO₄)₃ = 
$\frac{1684 g}{342 g mo{l}^{-1}}=4.92 mol [Molar mass of A{l}_2(S{O}_4{)}_3 =342 g mo{l}^{-1}]$

Chemical reaction:    $A{l}_2(S{O}_4{)}_3 + 3BaC{l}_2 \to 3BaS{O}_4 + 2 AlC{l}_3$

For 1 mol of Al₂(SO₄)₃ we will have 3 mol of BaSO4, therefore, for 4.92 mol of Al₂(SO₄)₃ it will produce 3 x 4.92 = 14.76 mol of BaSO₄.

Best Wishes !