Question

Calculate the molality of 1 litre solution of 93% H₂SO₄ (weight/volume). The density of the solution is 1.84 g mL⁻¹.

Answer 1

Given: Volume of solution = 1 litre = 1000 mL

Strength of H₂SO₄ = 93% w/v

⇒ 93 g of H₂SO₄ in 100 mL solution

Density of solution = 1.84 g/mL

Molar mass of H₂SO₄ = 98 g/mol

Mass of 1000 mL solution = 1000 × 1.84 = 1840 g

Since it is 93% w/v,

In 100 mL, 93 g H₂SO₄

So in 1000 mL,

Solute = 93 × 10 =  930 g

Mass of solvent = Mass of solution − Mass of solute

= 1840 − 930

= 910 g

Mass of solvent = 0.910 kg

Moles = `930/98` = 9.49 mol

Molality (m) = `"Moles of solute"/"Mass of solvent (kg)"`

Molality (m) = `9.49/0.910`

Molality (m) = 10.43 mol/kg

∴ The molality of 1 litre solution is 10.43 mol/kg.