CLASS 11–12 CHEMISTRY · SOLUTIONS

Molarity, molality and dilution

Molarity M = moles of solute ÷ volume of solution in litres. Molality m = moles of solute ÷ mass of solvent in kg. On dilution, M₁V₁ = M₂V₂.

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Worked examples

Example 1 · level 1

Ayaan dissolves 3.725 g of potassium chloride (KCl) in water and makes the solution up to 100 mL. What is the molarity of the solution, in mol/L? (Atomic masses: K = 39, Cl = 35.5)

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M(KCl) = 39 + 35.5 = 74.5 g/mol.
n = 3.725 ÷ 74.5 = 0.05 mol.
V = 100 mL = 0.1 L.
Molarity = 0.05 ÷ 0.1 = 0.5 mol/L.

Answer: 0.5

Example 2 · level 2

4 g of sodium hydroxide (NaOH) is dissolved in 250 g of water. What is the molality of the solution, in mol/kg? (Atomic masses: Na = 23, O = 16, H = 1)

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M(NaOH) = 23 + 16 + 1 = 40 g/mol.
n = 4 ÷ 40 = 0.1 mol.
Mass of water = 250 g = 0.25 kg.
Molality = 0.1 ÷ 0.25 = 0.4 mol/kg.

Answer: 0.4

Example 3 · level 3

A school lab in Bhopal has a 2 M stock solution of HCl. Aarav needs 1000 mL of 0.25 M HCl. What volume of the stock solution (in mL) must be diluted with water? Give the answer correct to 2 decimal places.

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M₁V₁ = M₂V₂ → 2 × V₁ = 0.25 × 1000.
V₁ = 250 ÷ 2 = 125 mL.
Take 125 mL of stock and make it up to 1000 mL with water.

Answer: 125 mL