Question

A solution of Al₂(SO₄)₃ {d=1.253 g / cm3} contain 40% salt by weight. The molarity is ​

Answer 1

A solution of Al₂(SO₄)₃ {d=1.253 gm / ml} contain 22% salt by weight.

means, 22g of Al₂(SO₄)₃ in 100g of solution.

molecular wt of Al₂(SO₄)₃ = 2 × 27 + 3 × 96 = 342g/mol

so, no of mole of Al₂(SO₄)₃ = 22g/342g/mol = 0.0643

volume of solution = mass/density

= 100g/1.253 g/ml

= 79.8ml

now molarity = no of mole × 1000/volume of solution

= 0.0643 × 1000/79.8

= 64.3/79.8

≈ 0.805 M

molality = 0.643 × 1000/mass of solvent

= 0.643 × 1000/(100g - 22g)

= 64.3/78

= 0.825 m

normality = n- factor × molarity

= 6 × 0.805

= 4.83 N