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IB DP Chemistry -S1.4.5 Molar concentration – Study Notes

IB DP Chemistry - S1.4.5 Molar concentration - Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – S1.4.5 Molar concentration – Study Notes – New Syllabus

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Structure 1.4.5 — Molar Concentration, Amount of Solute, and Volume of Solution

Structure 1.4.5 — Molar Concentration, Amount of Solute, and Volume of Solution

 Molar Concentration

Molar concentration, also known as molarity, refers to the number of moles of solute dissolved in one cubic decimetre (1 dm³) of solution.

 

It provides a quantitative measure of the amount of substance present in a given volume of solution.

Represented as: [X], where X is the chemical formula or symbol of the solute.

Example: [NaCl] = 0.25 mol dm⁻³ means 0.25 mol of NaCl is dissolved per dm³ of solution.

 Formula (from the data booklet):

\( n = C \times V \)

  • \( n \) = amount of solute in moles (mol)
  • \( C \) = molar concentration in mol dm⁻³
  • \( V \) = volume of solution in dm³

Rearranged Forms:

  • \( C = \frac{n}{V} \)
  • \( V = \frac{n}{C} \)

Unit Conversions:

  • Volume: cm³ → dm³ → divide by 1000
  • Mass concentration: g dm⁻³
  • To convert molar concentration to mass concentration:
    \( C_{\text{g dm}^{-3}} = C_{\text{mol dm}^{-3}} \times M \) (M = molar mass in g mol⁻¹)
  • To convert mass concentration to molar concentration:
    \( C_{\text{mol dm}^{-3}} = \frac{C_{\text{g dm}^{-3}}}{M} \)

Additional Notes:

  • Ensure all volumes are expressed in dm³ before substituting into the formula.
  • Concentration is a temperature-dependent property – use standard conditions unless stated otherwise.
  • Use a calibrated volumetric flask to prepare solutions of known concentration accurately.
  • Label all concentrations clearly using square brackets and correct units (e.g., [HCl] = 2.0 mol dm⁻³).

Example

Calculate the concentration in mol dm⁻³ of a solution containing 4.00 g of NaOH in 250 cm³ of solution.

▶️ Answer/Explanation
  • Molar mass of NaOH = 23.00 + 16.00 + 1.008 = 40.01 g mol⁻¹
  • Moles of NaOH = \( \frac{4.00}{40.01} = 0.100 \, \text{mol} \)
  • Volume = 250 cm³ = 0.250 dm³
  • \( C = \frac{n}{V} = \frac{0.100}{0.250} = 0.400 \, \text{mol dm}^{-3} \)

Example

A 0.200 mol dm⁻³ HCl solution is prepared by dissolving a certain mass of HCl in 500 cm³ of water. Calculate the mass of HCl used.

▶️ Answer/Explanation
  • Volume = 500 cm³ = 0.500 dm³
  • Moles = \( C \times V = 0.200 \times 0.500 = 0.100 \, \text{mol} \)
  • Molar mass of HCl = 1.008 + 35.45 = 36.46 g mol⁻¹
  • Mass = \( n \times M = 0.100 \times 36.46 = 3.65 \, \text{g} \)

Example

25.0 cm³ of a sulfuric acid solution was completely neutralised by 35.0 cm³ of 0.150 mol dm⁻³ sodium hydroxide. Calculate the concentration of the sulfuric acid solution in mol dm⁻³.

(Equation: \( \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \))

▶️ Answer/Explanation
    • Step 1 – Calculate moles of NaOH:
    • Volume = 35.0 cm³ = 0.0350 dm³
    • \( n(\text{NaOH}) = C \times V = 0.150 \times 0.0350 = 0.00525 \, \text{mol} \)
    • Step 2 – Use stoichiometry:
    • From the balanced equation: 2 mol NaOH reacts with 1 mol H₂SO₄
    • \( n(\text{H}_2\text{SO}_4) = \frac{0.00525}{2} = 0.002625 \, \text{mol} \)
  • Step 3 – Calculate concentration of H₂SO₄:
  • Volume = 25.0 cm³ = 0.0250 dm³
  • \( C = \frac{n}{V} = \frac{0.002625}{0.0250} = 0.105 \, \text{mol dm}^{-3} \)

Answer: 0.105 mol dm⁻³

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