Concentration in mol/dm³ & gas volumes (Chem only, HT) – GCSE Chemistry exam questions with mark scheme

1Grade 8–9HigherTriple only5 marks

A student titrated 25.0 cm³ of sodium hydroxide solution with hydrochloric acid of concentration 0.100 mol/dm³.

$$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

	Rough	Run 1	Run 2	Run 3
Volume of acid in cm³	21.50	20.80	20.90	20.85

(a) Calculate the concentration of the sodium hydroxide solution in mol/dm³. Use the concordant results. Give your answer to 3 significant figures. [4]

(b) Convert your answer to (a) into g/dm³. $M_r$ of NaOH = 40 [1]

(a) mean titre $= (20.80 + 20.90 + 20.85) \div 3 = 20.85\ \text{cm}^3$ (rough excluded)(1)
(a) moles HCl $= 0.100 \times \dfrac{20.85}{1000} = 2.085 \times 10^{-3}$(1)
(a) moles NaOH $= 2.085 \times 10^{-3}$ (1 : 1 ratio)(1)
(a) concentration $= 2.085 \times 10^{-3} \div 0.0250 = 0.0834\ \text{mol/dm}^3$(1)
(b) $0.0834 \times 40 = 3.34\ \text{g/dm}^3$ allow ecf from (a); allow 3.3(1)
Answer: (a) 0.0834 mol/dm³ (b) 3.34 g/dm³
Examiner tip: Leave out the rough titre and divide by 0.0250 dm³, not 25.0.

2Grade 4–5HigherTriple only4 marks

A technician makes a solution of potassium hydroxide, KOH, with a concentration of 0.250 mol/dm³.

A student measures out 40.0 cm³ of this solution.

(a) Calculate the amount, in moles, of potassium hydroxide in 40.0 cm³ of the solution. [2]

(b) Calculate the mass of potassium hydroxide in 40.0 cm³ of the solution.

Relative formula mass $(M_r)$ of KOH = 56 [1]

(a) volume $= 40.0 \div 1000 = 0.0400$ dm³(1)
(a) moles $= 0.250 \times 0.0400 = 0.0100$ (mol) allow 0.01; correct answer with no working scores 2(1)
(b) mass $= 0.0100 \times 56 = 0.56$ (g) allow ecf from (a)(1)
(c) $0.250 \times 56 = 14$ (g/dm³) allow $0.56 \div 0.0400 = 14$ allow ecf from (b)(1)
Answer: (a) 0.0100 mol (b) 0.56 g (c) 14 g/dm³
Examiner tip: Divide 40.0 cm³ by 1000 before multiplying: 0.250 × 40.0 = 10 mol is the classic slip and costs both marks in (a).

3Grade 4–5HigherTriple only4 marks

Solid carbon dioxide (dry ice) changes directly into carbon dioxide gas at room temperature.

(a) Give the volume occupied by one mole of any gas at room temperature and pressure (RTP). Give the unit. [1]

(b) A piece of dry ice has a mass of 8.8 g.

Calculate the volume of carbon dioxide gas produced at RTP when the piece of dry ice changes into a gas.

Relative formula mass $(M_r)$ of CO₂ = 44 [2]

(a) 24 dm³ allow 24 000 cm³; unit required(1)
(b) moles $= 8.8 \div 44 = 0.20$ (mol)(1)
(b) volume $= 0.20 \times 24 = 4.8$ dm³ allow 4800 cm³; unit required for this mark; allow ecf for moles × 24(1)
(c) 4.8 dm³ allow 4800 cm³; allow 0.20 × their value from (a); do not accept ‘the same volume as the CO₂’ without a volume(1)
Answer: (a) 24 dm³ (b) 4.8 dm³ (c) 4.8 dm³
Examiner tip: Moles of any gas, not its mass, decide its volume: 0.20 mol of N₂ takes up exactly the same 4.8 dm³ as 0.20 mol of CO₂.

4Grade 6–7HigherTriple only4 marks

Propane is used as a fuel in camping stoves. It burns completely in oxygen.

$$\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$$

All gas volumes are measured at room temperature and pressure.

(a) Calculate the volume of oxygen needed to burn 60 cm³ of propane completely. [1]

(b) Calculate the volume of carbon dioxide produced when 60 cm³ of propane burns completely. [1]

(c) A student mixes 60 cm³ of propane with 400 cm³ of oxygen and burns the mixture completely. The water produced is a liquid.

Calculate the total volume of gas left at the end. [2]

(a) 300 cm³(1)
(b) 180 cm³(1)
(c) oxygen left over $= 400 - 300 = 100$ cm³ allow ecf from (a)(1)
(c) total $= 100 + 180 = 280$ cm³ allow ecf from (b) and from the first mark; do not credit an answer that includes a volume for water (e.g. 520 cm³)(1)
Answer: (a) 300 cm³ (b) 180 cm³ (c) 280 cm³ (100 cm³ O₂ + 180 cm³ CO₂)
Examiner tip: Gas volumes are in the same ratio as the moles in the equation, so no conversion to moles is needed; just leave out the water because it is a liquid at RTP.

5Grade 6–7HigherTriple only5 marks

A student makes 250 cm³ of sodium carbonate solution with a concentration of 0.100 mol/dm³ in a volumetric flask.

(a) Calculate the mass of sodium carbonate the student must dissolve.

Relative formula mass $(M_r)$ of Na₂CO₃ = 106 [2]

(b) The student uses a pipette to transfer 25.0 cm³ of this solution into a clean 100 cm³ volumetric flask. The student then adds distilled water up to the 100 cm³ mark and mixes.

Calculate the concentration of the new solution in mol/dm³. [2]

Explain why. [1]

(a) moles $= 0.100 \times \dfrac{250}{1000} = 0.0250$ (mol)(1)
(a) mass $= 0.0250 \times 106 = 2.65$ (g) allow 2.7; correct answer with no working scores 2(1)
(b) moles in 25.0 cm³ $= 0.100 \times 0.0250 = 0.00250$ (mol) or volume has increased by a factor of 4(1)
(b) concentration $= 0.00250 \div 0.100 = 0.0250$ (mol/dm³) allow 0.025; correct answer with no working scores 2(1)
(c) the same amount (number of moles) of sodium carbonate is now in a larger volume (of solution) allow same number of particles spread through more water; ignore 'it is diluted' alone(1)
Answer: (a) 2.65 g (b) 0.0250 mol/dm³ (c) same number of moles in a larger volume
Examiner tip: In (b) the moles come from the 25.0 cm³ that was moved, not the whole 250 cm³: dividing 0.0250 mol by 0.100 dm³ gives 0.250, ten times too big.

6Grade 8–9HigherTriple only5 marks

A student used a titration to find the concentration of a sample of dilute sulfuric acid.

25.0 cm³ of potassium hydroxide solution of concentration 0.120 mol/dm³ was put into a conical flask.
The sulfuric acid was added from a burette until the indicator changed colour.

$$\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

	Rough	1	2	3
Final burette reading in cm³	15.70	15.55	30.50	16.60
Initial burette reading in cm³	0.00	0.50	15.55	1.20

(a) Concordant results are within 0.10 cm³ of each other.

Calculate the mean volume of sulfuric acid used. Use only the concordant results. [2]

(b) Calculate the concentration of the sulfuric acid in mol/dm³. [3]

(a) titres 15.70, 15.05, 14.95, 15.40 (cm³) and titrations 1 and 2 identified as concordant may be implied by the correct mean(1)
(a) mean $= (15.05 + 14.95) \div 2 = 15.00$ (cm³) allow 15.0 or 15(1)
(b) moles KOH $= 0.120 \times \dfrac{25.0}{1000} = 0.00300$ (mol)(1)
(b) moles H₂SO₄ $= 0.00300 \div 2 = 0.00150$ (mol) allow ecf from moles of KOH(1)
(b) concentration $= 0.00150 \div \dfrac{15.00}{1000} = 0.100$ (mol/dm³) allow 0.1; allow ecf from (a) and from the moles of acid; an answer of 0.200 (ratio not used) scores 2(1)
Answer: (a) 15.00 cm³ (b) 0.100 mol/dm³
Examiner tip: Check the equation before dividing: 2 KOH react with 1 H₂SO₄, so the moles of acid are half the moles of alkali. Missing this gives 0.200 mol/dm³.

7Grade 8–9HigherTriple only6 marks

A student investigated the reaction between magnesium and excess dilute hydrochloric acid.

$$\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

The student reacted different masses of magnesium ribbon and collected the hydrogen in a gas syringe. All volumes were measured at room temperature and pressure (RTP).

The graph shows the results.

(a) One result is anomalous.

Give the mass of magnesium used for the anomalous result and suggest one reason why the volume of hydrogen was too low. [2]

(b) Use the graph to determine the volume of hydrogen produced by 0.045 g of magnesium. [1]

(a) 0.030 g(1)
(a) any one from: (some) hydrogen escaped before the bung was put in / the bung was not put in quickly enough; leak in the apparatus / gas syringe not sealed; the syringe was read before the reaction had finished; less magnesium was used than recorded ignore 'human error' or 'measured wrongly' alone(1)
(b) 45 cm³ allow 44–46 cm³(1)
(c) moles H₂ $= 45 \div 24\,000 = 0.001875$ (mol) allow $0.045 \div 24$; allow ecf from (b)(1)
(c) moles Mg = moles H₂ $= 0.001875$ (1 : 1 ratio from the equation)(1)
(c) $A_r = 0.045 \div 0.001875 = 24$ allow 23.5–24.5 from a reading in the range in (b); allow ecf; correct answer with no working scores 1 only(1)
Answer: (a) 0.030 g; e.g. some hydrogen escaped before the bung was fitted (b) 45 cm³ (c) $A_r = 24$
Examiner tip: Change cm³ to dm³ (or use 24 000 cm³) before dividing by the molar gas volume: 45 ÷ 24 gives a mole value 1000 times too large and an $A_r$ of 0.024.

8Grade 8–9HigherTriple only6 marks

A student added 0.65 g of zinc powder to 20.0 cm³ of dilute sulfuric acid of concentration 0.400 mol/dm³.

$$\text{Zn(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{H}_2\text{(g)}$$

Relative atomic mass $(A_r)$: Zn = 65

(a) Calculate the maximum volume of hydrogen gas, in cm³, that could be produced at room temperature and pressure (RTP).

You must show which reactant is the limiting reactant. [5]

(b) Calculate the mass of zinc left over when the reaction has finished. [1]

(a) moles Zn $= 0.65 \div 65 = 0.0100$ (mol)(1)
(a) moles H₂SO₄ $= 0.400 \times \dfrac{20.0}{1000} = 0.00800$ (mol)(1)
(a) sulfuric acid is the limiting reactant because fewer moles react in a 1 : 1 ratio or zinc is in excess must be supported by the two mole values(1)
(a) moles H₂ $= 0.00800$ (mol) allow ecf from an incorrectly identified limiting reactant for this mark and the next(1)
(a) volume $= 0.00800 \times 24\,000 = 192$ (cm³) allow 0.192 dm³ if unit given; an answer of 240 cm³ (zinc used as limiting) scores 4 if all other working is shown(1)
(b) $(0.0100 - 0.00800) \times 65 = 0.13$ (g) allow ecf from mole values in (a)(1)
Answer: (a) 192 cm³ (sulfuric acid is limiting) (b) 0.13 g
Examiner tip: Always work out the moles of BOTH reactants before choosing which one sets the amount of product: using the zinc gives 240 cm³ and loses the limiting-reactant mark.

9Grade 6–7HigherTriple only6 marks

A student is given potassium hydroxide solution of unknown concentration and nitric acid of concentration 0.100 mol/dm³.

$$\text{KOH} + \text{HNO}_3 \rightarrow \text{KNO}_3 + \text{H}_2\text{O}$$

Describe how the student should carry out a titration to find the volume of nitric acid that reacts exactly with 25.0 cm³ of the potassium hydroxide solution.

Explain how the student should then use the results to calculate the concentration of the potassium hydroxide solution in mol/dm³. [6]

use a (volumetric) pipette (and pipette filler) to measure 25.0 cm³ of potassium hydroxide solution into a conical flask(1)
add a few drops of a suitable indicator, e.g. phenolphthalein or methyl orange do not accept universal indicator(1)
fill a burette with the nitric acid (using a funnel, then remove the funnel) and record the initial reading(1)
place the flask on a white tile; add the acid from the burette while swirling the flask(1)
add the acid drop by drop near the end-point(1)
stop when the indicator just changes colour (permanently), e.g. phenolphthalein pink to colourless; record the final reading(1)
volume of acid used = final reading − initial reading; read the burette at eye level / bottom of the meniscus(1)
do a rough titration first, then repeat until concordant results (within 0.10 cm³) are obtained; calculate the mean of the concordant titres(1)
moles HNO₃ = 0.100 × (mean titre ÷ 1000)(1)
moles KOH = moles HNO₃ because the reacting ratio is 1 : 1(1)
concentration of KOH = moles KOH ÷ 0.0250 (dm³)(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that would give accurate, repeatable results, naming the correct apparatus, the use of an indicator, how the end-point is found and repeating to concordant results. The calculation is explained in full: moles of acid, use of the 1 : 1 ratio and division by the volume of alkali in dm³.
Level 2	3–4	A method that would work, with the main apparatus and an indicator, but some steps that improve accuracy are missing or unclear. Some correct steps of the calculation are given, or the method is detailed but the calculation is missing.
Level 1	1–2	Some relevant steps or apparatus are given, but the method is not in a logical order and would not reliably find the reacting volume. Little or no correct calculation.
No relevant content	0	Nothing worthy of credit.

Answer: Pipette 25.0 cm³ KOH into a conical flask with indicator; add 0.100 mol/dm³ HNO₃ from a burette, swirling, dropwise near the end-point, until the colour just changes; repeat to concordant titres and take the mean. Then moles HNO₃ = 0.100 × titre/1000 = moles KOH (1 : 1); concentration = moles ÷ 0.0250.
Examiner tip: Top answers say how the 25.0 cm³ is measured (pipette, not a measuring cylinder), how the end-point is judged, and finish with all three calculation steps including dividing by 0.0250 dm³.

10Grade 6–7HigherTriple only5 marks

A teacher has a sample of an unknown gas. The sample has a mass of 1.60 g and a volume of 600 cm³ at room temperature and pressure (RTP).

The gas is one of those in the table.

Gas	Formula	$M_r$
ammonia	NH₃	17
oxygen	O₂	32
sulfur dioxide	SO₂	64
chlorine	Cl₂	71

(a) Identify the gas. You must show your working. [3]

(b) A second sample contains 600 cm³ of oxygen at RTP.

Compare the number of molecules and the mass of this oxygen sample with the sample of the unknown gas. [2]

(a) moles $= 600 \div 24\,000 = 0.0250$ (mol) allow $0.600 \div 24$(1)
(a) $M_r = 1.60 \div 0.0250 = 64$(1)
(a) sulfur dioxide / SO₂ allow ecf from a calculated $M_r$ that matches a gas in the table; no mark for SO₂ with no working(1)
(b) same number of molecules (because equal volumes of gases at the same temperature and pressure contain the same amount in moles)(1)
(b) the oxygen has a smaller mass: half the mass / 0.80 g allow ecf from (a) for 0.0250 × 32; must give a comparison, not just a mass(1)
Answer: (a) 0.0250 mol, $M_r$ = 64, sulfur dioxide (b) same number of molecules; oxygen has half the mass (0.80 g)
Examiner tip: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, not equal masses: the heavier molecules make the SO₂ sample twice the mass.

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