Dabour Hub EGEgyptian Curriculum 2026/2027
1 Molarity

1 Molarity

Point 17 of 20 — Chapter 1 — Chemical CalculationQuestion bank: 2 questions (0 with pictures)

17 6.1 Molarity

  • The concentration of a solution can be expressed in several ways.
  • The most important is molar concentration (molarity): the number of moles of solute dissolved in one litre of solution.

Molarity M (mol/L) = number of moles of solute n (mol) / volume of solution V (L)

Units: mol/L, written M. If the volume is given in mL or cm³, divide by 1000. Also n = M × V (L), mass of solute = M × V (L) × molar mass. Note: V is the volume of the solution, not of the solvent.

Worked example 6.1. Molarity of a solution made by dissolving 4.9 g of sulphuric acid in water to make 250 mL. [H = 1, S = 32, O = 16]

  • M(H₂SO₄) = 98; n = 4.9 / 98 = 0.05 mol; V = 0.25 L.
  • M = 0.05 / 0.25 = 0.2 mol/L.

Worked example 6.2 — mass needed. Mass of Na₂CO₃ needed to prepare 500 mL of 0.2 M solution [Na = 23, C = 12, O = 16]. n = 0.2 × 0.5 = 0.1 mol; mass = 0.1 × 106 = 10.6 g.

Worked example 6.3 — pattern of the NaOH example. 2 g of NaOH (40) in 500 mL: n = 0.05 mol; M = 0.05 / 0.5 = 0.1 M.

📝 Question bank for this point

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