Volume from Concentration Calculations

Finding the volume of a solution when you know its concentration and the number of moles of solute is a direct rearrangement of the concentration formula. This type of question appears regularly in IGCSE Chemistry, both as standalone problems and as steps within larger titration calculations.

The method

1. Identify the concentration (c) in mol/dm3 and the number of moles (n).
2. If moles are not given directly, calculate them from mass using n = mass / Mr.
3. Apply the formula V = n / c to find the volume in dm3.
4. Convert to cm3 if the question requires it (multiply by 1000).

The formula

V = n / c

Where:

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

Worked examples

Example 1: Simple volume calculation

Question: What volume of 0.5 mol/dm3 hydrochloric acid contains 0.1 mol of HCl?

Working:

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

Answer: 200 cm3

Examiner note: Straightforward substitution. Always state whether your final answer is in dm3 or cm3.

Example 2: Volume from mass and concentration

Question: What volume of 0.2 mol/dm3 sodium hydroxide solution is needed to dissolve 2 g of NaOH? (Ar: Na = 23, O = 16, H = 1)

Working:

• Mr of NaOH = 23 + 16 + 1 = 40
• Moles of NaOH = 2 / 40 = 0.05 mol
• V = n / c = 0.05 / 0.2 = 0.25 dm3
• Convert: 0.25 x 1000 = 250 cm3

Answer: 250 cm3

Examiner note: This is a two-step problem. Show the moles calculation before the volume calculation.

Example 3: Volume of acid for a reaction

Question: What volume of 2 mol/dm3 hydrochloric acid is needed to react with 0.1 mol of sodium carbonate? The equation is Na2CO3 + 2HCl -> 2NaCl + H2O + CO2.

Working:

• Mole ratio: 1 Na2CO3 : 2 HCl
• Moles of HCl needed = 0.1 x 2 = 0.2 mol
• V = n / c = 0.2 / 2 = 0.1 dm3
• Convert: 0.1 x 1000 = 100 cm3

Answer: 100 cm3

Examiner note: The mole ratio step is critical. Without it, you would calculate the volume for 0.1 mol of HCl instead of 0.2 mol.

Example 4: Small volumes

Question: What volume of 0.1 mol/dm3 silver nitrate solution contains 0.005 mol of AgNO3?

Working:

• V = n / c = 0.005 / 0.1 = 0.05 dm3
• Convert: 0.05 x 1000 = 50 cm3

Answer: 50 cm3

Examiner note: Small volumes are common in titration-style questions. Keep your working in dm3 until the final conversion.

Example 5: Finding volume when given mass of a different substance

Question: What volume of 0.5 mol/dm3 sulfuric acid is needed to react completely with 4 g of magnesium? (Ar: Mg = 24) The equation is Mg + H2SO4 -> MgSO4 + H2.

Working:

• Moles of Mg = 4 / 24 = 0.167 mol (3 s.f.)
• Mole ratio: 1 Mg : 1 H2SO4
• Moles of H2SO4 = 0.167 mol
• V = n / c = 0.167 / 0.5 = 0.333 dm3
• Convert: 0.333 x 1000 = 333 cm3

Answer: 333 cm3

Examiner note: This combines reacting masses logic with the volume formula. The sequence is: mass -> moles -> mole ratio -> moles -> volume.

Common mistakes

1. Forgetting to apply the mole ratio. When the question involves a chemical reaction, you must use the balanced equation to convert between moles of different substances before finding volume.
2. Mixing up dm3 and cm3. Remember that 1 dm3 = 1000 cm3. Check which unit the question requires.
3. Using mass instead of moles. The formula V = n / c requires moles. If you have mass, convert first.
4. Dividing concentration by moles. The formula is V = n / c, not V = c / n. Write the triangle (n at top, c and V at the bottom) if it helps you remember.

When this appears

Volume from concentration questions appear on Paper 2 and Paper 4. This calculation is a key step in titration calculations and works alongside concentration calculations. See the stoichiometry topic for the broader context.