Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

1.4.4Additional lessons

Molar Concentration — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 17 of 172

📚 IB Chemistry HL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
Curriculum Equation Key قوانين المول والحسابات الكيميائية - Moles & Stoichiometry FormulasFast Exam Solution
n = m / M_r | N = n × N_A (6.022 × 10²³) | M = n / V | V_gas = n × 22.4 L (STP)
Word Equation & Conditions: عدد المولات = الكتلة بالجرام ÷ الكتلة المولية | عدد الجسيمات = عدد المولات × عدد أفوجادرو
Location & Specifications: كيمياء التحصيلي وثانوي وبكالوريا و Cambridge IGCSE Chemistry.
0%read

Section 1 of 5Concentration

The concentration of a solution tells us how much solute is dissolved in a given volume of solution.

Definition
  • Definition Molar concentration c = n ÷ V, where n is the amount of solute in mol and V is the volume in dm³.
Units
  • Units The unit is mol dm⁻³, sometimes written M, and the square-bracket notation [X] means the concentration of X.
Conversion
  • Conversion 1 dm³ = 1000 cm³, so volume in cm³ must be divided by 1000.
Mass concentration
  • Mass concentration Concentration may also be given in g dm⁻³, which is mass of solute divided by volume.

Section 2 of 5Calculation Steps

Most problems can be solved with these steps.

1
Convert the volume from cm³ to dm³ by dividing by 1000.
2
If the mass of solute is given, divide it by the molar mass to find the amount in moles.
3
Calculate concentration using c = n ÷ V, or rearrange to n = c × V.
4
For dilution, the amount of solute stays constant, so c₁V₁ = c₂V₂.
5
Check that the answer has the correct unit and sensible size.

Section 3 of 5Related Ideas

Related Ideas
Parts per million
  • Parts per million ppm is mass of solute per million parts of mass of solution, used for very dilute solutions.
Standard solution
  • Standard solution A solution of accurately known concentration made in a volumetric flask.
Titration
  • Titration The known concentration of a standard solution is used to find an unknown concentration.
Ions
  • Ions For Na₂SO₄ of 0.10 mol dm⁻³, [Na⁺] = 0.20 mol dm⁻³ and [SO₄²⁻] = 0.10 mol dm⁻³.

Section 4 of 5Worked Example

Question

  • 4.00 g of NaOH is dissolved to make 250 cm³ of solution.
  • Find the concentration, then the concentration after diluting 25.0 cm³ to 100 cm³.

  1. M(NaOH) = 40.00 g mol⁻¹, so n = 4.00 ÷ 40.00 = 0.100 mol.
  2. c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³.
  3. c₂ = 0.400 × 25.0 ÷ 100 = 0.100 mol dm⁻³.

Answer

  • 0.400 mol dm⁻³;
    • after dilution 0.100 mol dm⁻³

Section 5 of 5Exam Checklist & Tips

1.4.4 – Key Points for the Exam

  • Concentration Amount of solute per unit volume of solution.
  • Solute The substance dissolved in a solvent.
  • Dilution Adding solvent to decrease concentration while the amount of solute stays the same.
  • Standard solution A solution whose concentration is known accurately.

Common Mistakes

  • Using the volume in cm³ without converting to dm³.
  • Using the volume of solvent rather than the total volume of solution.
  • Forgetting to count both ions when an ionic compound dissolves.
  • Stating that the moles of solute change on dilution.

Last updated:

\n