This set covers five skills: converting cm³ to dm³, using concentration, relating gas volume to moles, calculating percentage yield and purity, and explaining assumptions. All numbers are invented for practice.
Work in order, because the questions get harder. Use Aᵣ values H = 1, C = 12, O = 16, Na = 23, Mg = 24, S = 32, Ca = 40, Cu = 64, Zn = 65, and take the molar volume of a gas as 24 dm³/mol at room temperature and pressure. Open each answer only after you have written your own working.
Questions
Q1 (easy). Convert 35.0 cm³ to dm³.
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Divide by 1000: 35.0 ÷ 1000 = 0.0350 dm³.
Q2. How many moles of solute are in 25.0 cm³ of a solution of concentration 0.400 mol/dm³?
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Volume = 25.0 ÷ 1000 = 0.0250 dm³. Moles = 0.400 × 0.0250 = 0.0100 mol.
Q3. 0.0300 mol of solute is dissolved to make 150 cm³ of solution. Find the concentration in mol/dm³.
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Volume = 150 ÷ 1000 = 0.150 dm³. Concentration = 0.0300 ÷ 0.150 = 0.200 mol/dm³.
Q4. A sulfuric acid solution, H₂SO₄, has concentration 0.250 mol/dm³. Give its concentration in g/dm³.
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Mᵣ = (2 × 1) + 32 + (4 × 16) = 98. Concentration = 0.250 × 98 = 24.5 g/dm³.
Q5. In NaOH + HCl → NaCl + H₂O, 20.0 cm³ of 0.150 mol/dm³ NaOH is exactly neutralised by 24.0 cm³ of HCl. Find the concentration of the HCl.
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Moles of NaOH = 0.150 × 0.0200 = 0.00300 mol. Ratio NaOH : HCl is 1 : 1, so HCl = 0.00300 mol. Volume of HCl = 0.0240 dm³. Concentration = 0.00300 ÷ 0.0240 = 0.125 mol/dm³.
Q6. In H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, 25.0 cm³ of 0.100 mol/dm³ NaOH is exactly neutralised by 20.0 cm³ of sulfuric acid. Find the concentration of the acid.
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Moles of NaOH = 0.100 × 0.0250 = 0.00250 mol. Ratio H₂SO₄ : NaOH is 1 : 2, so H₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol. Concentration = 0.00125 ÷ 0.0200 = 0.0625 mol/dm³.
Q7. Zn + 2HCl → ZnCl₂ + H₂. A 1.30 g sample of zinc reacts completely with excess acid. Find the volume of hydrogen in cm³.
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Moles of Zn = 1.30 ÷ 65 = 0.0200 mol. Ratio Zn : H₂ is 1 : 1, so H₂ = 0.0200 mol. Volume = 0.0200 × 24 = 0.480 dm³ = 480 cm³.
Q8. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. What mass of calcium carbonate gives 240 cm³ of carbon dioxide?
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Moles of CO₂ = 0.240 ÷ 24 = 0.0100 mol. Ratio CaCO₃ : CO₂ is 1 : 1, so CaCO₃ = 0.0100 mol. Mᵣ = 40 + 12 + 48 = 100. Mass = 0.0100 × 100 = 1.00 g.
Q9. CuO + H₂ → Cu + H₂O. From 4.00 g of copper(II) oxide, a student obtains 2.56 g of copper. Find the percentage yield.
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Mᵣ of CuO = 64 + 16 = 80. Moles = 4.00 ÷ 80 = 0.0500 mol. Ratio CuO : Cu is 1 : 1, so Cu = 0.0500 mol. Theoretical mass = 0.0500 × 64 = 3.20 g. Yield = 2.56 ÷ 3.20 × 100 = 80.0%.
Q10. The theoretical mass of a product is 8.00 g. The solid collected weighs 7.20 g and is 95% pure. Find (a) the mass of pure product, and (b) the true percentage yield.
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(a) Pure mass = 7.20 × 0.95 = 6.84 g.
(b) True yield = 6.84 ÷ 8.00 × 100 = 85.5%.
The apparent yield from the whole sample would be 7.20 ÷ 8.00 × 100 = 90%, which overstates the result.
Q11 (harder). Mg + 2HCl → MgCl₂ + H₂. A student has 2.40 g of magnesium and 0.150 mol of hydrochloric acid. Find the limiting reactant and the volume of hydrogen formed, in dm³.
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Moles of Mg = 2.40 ÷ 24 = 0.100 mol. All the magnesium needs 2 × 0.100 = 0.200 mol of HCl, but only 0.150 mol is present. HCl is limiting.
Moles of H₂ = 0.150 ÷ 2 = 0.0750 mol. Volume = 0.0750 × 24 = 1.80 dm³.
Q12 (harder). A calculation predicts 0.240 dm³ of oxygen from a decomposition reaction, but 0.198 dm³ is measured. Give two different assumptions that may not have held and state the effect of each.
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The measured value is 0.198 ÷ 0.240 × 100 = 82.5% of the prediction. Two valid answers:
- The reaction may not have gone to completion, so less reactant decomposed and less gas formed than predicted.
- Some gas may have escaped before it was measured, so the measured volume is lower than predicted.
A third option is that the starting material was impure, so there was less of the named substance than the mass suggests.
If you got these wrong
- Q1 to Q4 or an answer 1000 times off: revisit converting solution volume before using concentration.
- Q7, Q8 or Q11 gas volumes wrong: check the ratio line and the molar volume in relating gas amount to volume.
- Q9 or Q10 percentage wrong: check which mass is the numerator and which the denominator in calculating a percentage yield.
- Q10 apparent versus true yield: see distinguishing yield from purity.
- Q11 or Q12 reasoning: see explaining an assumption behind a calculation.
If the ratio or atomic masses are the problem, go back to the relative masses and amounts module. The mole and equation-ratio tutor, the equation balance reasoning trainer and the mistake log and retest queue can help you check steps and record errors for a later retest.
The module overview shows where each lesson sits. If you keep making the same slip after revising, a teacher in online one-to-one Chemistry tuition can watch your working line by line and spot it as it happens.