A solution of Al₂(SO₄)₃ {d=1.253 gm / ml} contain 22% salt by weight. The molarity, normality and molality of the solution is
(a) 0.805 M, 4.83 N, 0.825 m
(b) 0.825 M, 48.3 N, 0.805 m
(c) 4.83 M, 4.83 N, 4.83 m
(d) None of these
Answers
answer : option (a) 0.805 M, 4.83 N , 0.825 m
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
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