Question

What is the molarity and molality of a 13% solution by weight of sulphurics acid with a density of 1.02g/ml. To waht volume should 100ml of this solution be diluted in order to prepare a 1.5N solution?

Please Explain in detail and as simple as you can. Thank you for helping.

Answer 1

(i) Molarity = Moles of solute/Volume of solution in L

(ii) N₁V₁ = N₂V₂

A 13% solution (by weight) contains 13 g of solute (i.e H₂SO₄) per 100 gm of solution

Moles of solute = Mass of H₂SO₄/M. wt. of H₂SO₄ = 13/98 = 0.1326

Volume of solution in L

= Mass of solution/density of solution *1000 = 100/1.02 * 1000 = 0.0980 Litre

∴ Molarity of solution = 0.1326/0.0980 = 1.35 M

Again, Molality = Moles of solute/Mass of solvent in kg

Mass of solute in 100 ml of solution = 13 g [13% solution]

Mass of solvent = Mass of solution – Mass of solvent

= 100 – 13 = 87 g

∴ Molality = 13/98/87/100 = 1.57 m

Normality = Miolarity * Mol. wt/Eq. wt. or 1.35 * 98/49 = 2.70 N

N₁ = 2.70, V₁ = 100 ml, N₂ = 1.5, V₂= ? [∵ Eq. wt = 98/2 2H₂SO₄ = 49]

N₁V₁ = N₂V₂

2.70 * 100 = 1.5 * v₂­

Or V₂ = 2.70 *100/1.5 = 180 ml.

∴ 100 ml of this acid should be diluted to 180 ml to prepare 1.5 solutions.