A bottle of commercial H2SO4 (density = 1.787 g/mL) is labelled as 86% by mass.
(a) What is the molarity of the acid?
(b) What volume of the acid has to be used to make 1 litre 0.2 M H2SO4?
(c) What is the molality of the acid?
Answers
Answer- The above question is from the chapter 'Some Basic Concepts of Chemistry'.
Molarity: Number of moles of solute present in 1 litre of solution is called molarity.
Unit of Molarity = Molar (M)
•When volume is given in litres,
Molarity = Number of moles of solute ÷ Volume of solution in litres
•When volume is given in millilitres
Molarity = (Number of moles of solute × 1000) ÷ Volume of solution in ml
•When mass % and density is given,
Molarity = (Mass % × Density × 10) ÷ Molar mass of Solute
Molality: It is the number of moles of solution in 1 kg of solvent.
Unit of Molality: Molal (m)
•When mass of solvent is in kilograms,
Molality = Moles of solute ÷ Mass of solvent in kilograms
•When mass of solvent is in grams,
Molality = Moles of solute × 1000 ÷ Mass of solvent in grams
Relation between Molarity, Molality and Density:
=
+
Given question: A bottle of commercial H₂SO₄ (density = 1.787 g/mL) is labelled as 86% by mass.
(a) What is the molarity of the acid?
(b) What volume of the acid has to be used to make 1 litre 0.2 M H2SO4?
(c) What is the molality of the acid?
Answer: Molar mass of H₂SO₄ = 98 g
Mass = 86 g in 100 g of solution
No. of moles = = 86/98 = 0.88 moles
Density = 1.787 g/mL
= 1.787 g/mL
Volume = = 56 mL
(a) Molarity = (Number of moles of solute × 1000) ÷ Volume of solution in ml
M = (0.88 × 1000) ÷ 56 = 880 ÷ 56 = 15.71
∴ Molarity = 15.71 M
(b) We know that M₁V₁ = M₂V₂
15.71 × V₁ = 0.2 × 1
V₁ = 0.2 ÷ 15.71
V₁ = 0.0127 = 0.013 L
∴ Volume required = 0.013 L
(c) Molality = Moles of solute × 1000 ÷ Mass of solvent in grams
m = (0.88 × 1000) ÷ 98 = 880 ÷ 98 = 8.98