Concentration Calculations

Concentration calculations link the amount of solute dissolved in a solution to the volume of that solution. At IGCSE level, you need to be able to find any one of the three variables (concentration, moles, volume) when given the other two.

The method

1. Identify which of the three variables the question asks you to find: concentration (c), moles (n), or volume (V).
2. Rearrange the formula to solve for the unknown.
3. Check that volume is in dm3. If it is given in cm3, divide by 1000.
4. Substitute the values and calculate.

The formula

c = n / V

Where:

• c is concentration in mol/dm3
• n is the number of moles of solute (mol)
• V is the volume of solution in dm3

Rearranged:

• n = c x V
• V = n / c

Worked examples

Example 1: Finding concentration

Question: 0.2 mol of sodium hydroxide is dissolved in 500 cm3 of solution. Calculate the concentration.

Working:

• Convert volume: 500 cm3 = 500 / 1000 = 0.5 dm3
• c = n / V = 0.2 / 0.5 = 0.4 mol/dm3

Answer: 0.4 mol/dm3

Examiner note: The most common error is forgetting to convert cm3 to dm3. Always check the units of volume before substituting.

Example 2: Finding moles from concentration and volume

Question: How many moles of hydrochloric acid are present in 250 cm3 of a 0.1 mol/dm3 solution?

Working:

• Convert volume: 250 cm3 = 0.25 dm3
• n = c x V = 0.1 x 0.25 = 0.025 mol

Answer: 0.025 mol

Examiner note: This rearrangement is used frequently in titration calculations where you need to find moles from a known concentration and volume.

Example 3: Finding volume

Question: What volume of 0.5 mol/dm3 sulfuric acid contains 0.1 mol of H2SO4?

Working:

• V = n / c = 0.1 / 0.5 = 0.2 dm3
• Convert to cm3 if required: 0.2 x 1000 = 200 cm3

Answer: 200 cm3 (or 0.2 dm3)

Examiner note: Check what units the question expects for the answer. If it asks for cm3, multiply your dm3 answer by 1000.

Example 4: Concentration in g/dm3

Question: 5.85 g of sodium chloride is dissolved in 500 cm3 of water. Calculate the concentration in g/dm3. (Ar: Na = 23, Cl = 35.5)

Working:

• Convert volume: 500 cm3 = 0.5 dm3
• Concentration in g/dm3 = mass / volume = 5.85 / 0.5 = 11.7 g/dm3

Answer: 11.7 g/dm3

Examiner note: When the question asks for g/dm3 rather than mol/dm3, use mass directly instead of converting to moles first.

Example 5: Converting between g/dm3 and mol/dm3

Question: A solution of sodium hydroxide has a concentration of 8 g/dm3. What is this in mol/dm3? (Ar: Na = 23, O = 16, H = 1)

Working:

• Mr of NaOH = 23 + 16 + 1 = 40
• Concentration in mol/dm3 = concentration in g/dm3 / Mr
• c = 8 / 40 = 0.2 mol/dm3

Answer: 0.2 mol/dm3

Examiner note: To convert g/dm3 to mol/dm3, divide by Mr. To convert mol/dm3 to g/dm3, multiply by Mr.

Example 6: Finding mass of solute needed

Question: What mass of copper sulfate (CuSO4) is needed to make 250 cm3 of a 0.2 mol/dm3 solution? (Ar: Cu = 64, S = 32, O = 16)

Working:

• Convert volume: 250 cm3 = 0.25 dm3
• Moles needed: n = c x V = 0.2 x 0.25 = 0.05 mol
• Mr of CuSO4 = 64 + 32 + (4 x 16) = 160
• Mass = moles x Mr = 0.05 x 160 = 8 g

Answer: 8 g

Examiner note: This is a multi-step problem that combines concentration calculations with mass from moles. Show each step clearly.

Common mistakes

1. Not converting cm3 to dm3. This is the single most frequent error. Always divide by 1000 when volume is given in cm3.
2. Confusing g/dm3 with mol/dm3. These are different units. Check which the question asks for.
3. Using mass instead of moles. The formula c = n / V requires moles, not grams. If you are given a mass, convert to moles first using n = mass / Mr.
4. Rounding too early. Keep intermediate values to at least 3 significant figures and round only the final answer.

When this appears

Concentration calculations appear on Paper 2 and Paper 4. They are essential for titration calculations and connect to volume from concentration. See the stoichiometry topic for the broader context.