Concentration of Solutions (AQA GCSE Chemistry)

Exam Questions

2 hours25 questions
1a
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1 mark

Use the conversion factors in Figure 1 to convert the volumes in parts (a) - (e):

Figure 1

converting-units-aqa-gcse-3-4e-q1a

1000 cm3 = _____ dm3

1b
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1 mark

2500 cm3 = _____ dm3

1c
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1 mark

50 cm3 = _____ dm3

1d
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1 mark

_____ cm3 = 3.2 dm3

1e
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1 mark

_____ cm3 = 0.75 dm3

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2a1 mark

Write the equation to calculate concentration in terms of moles and volume.

2b
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2 marks

How many moles of sodium hydroxide are present in 1.5 dm3 of a 0.25 mol dm-3 sodium hydroxide solution?

2c
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1 mark

What is the concentration in g dm-3 of the 0.25 mol dm-3 solution of sodium hydroxide?

Relative formula mass (Mr):     NaOH = 40

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3a2 marks

In a titration experiment, 25.0 cm3 of dilute hydrochloric acid, HCl (aq), was neutralised by 23.5 cm3 of 0.100 mol dm-3 sodium hydroxide solution, NaOH (aq) to produce sodium chloride (NaCl) and water.

hydrochloric acid + sodium hydroxide → sodium chloride + water

Write the balanced symbol equation, including state symbols, for this reaction.

3b
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1 mark

Convert 23.5 cm3 into dm3.

3c
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2 marks

Calculate the number of moles of sodium hydroxide used in the titration.

3d
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1 mark

State the number of moles of hydrochloric acid neutralised by the sodium hydroxide.

3e
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2 marks

Calculate the concentration of the dilute hydrochloric acid.

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4a2 marks

A student neutralised dilute sulfuric acid with potassium hydroxide solution.

Balance the chemical equation for this reaction. 

___ KOH + ___ H2SO4 →  ___ K2SO4 + ___ H2O

4b
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3 marks

The student found that 31.0 cm3 of 0.250 mol / dm3 dilute sulfuric acid completely reacted with 25.0 cm3 of potassium hydroxide solution.

How many moles of sulfuric acid are used in the reaction?

4c
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3 marks

Calculate the concentration of the potassium hydroxide solution in mol / dm3.

4d
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2 marks

Calculate the concentration of the potassium hydroxide solution in g / dm3. 

Relative atomic masses (Ar):     H = 1;     K = 39;     O = 16

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5a
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2 marks

A technician needs to make a 0.150 mol / dm3 solution of barium chloride, BaCl2. 

What mass of barium chloride does the technician need to weigh out? 

Relative atomic masses (Ar):     Ba = 137;     Cl = 35.5

5b
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2 marks

The technician only has 250 cm3 volumetric flasks available to make the barium chloride solution.

What mass of barium chloride should the technician use to make the same concentration solution in a 250 cm3 flask?

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1
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1 mark

A solution of sulfuric acid, H2SO4, contains 9.8 g of the acid in 250 cm3 of solution.

What is the concentration in mol/dm3?

Relative atomic masses (Ar):   H = 1   S = 32   O= 16

  • 0.10 mol/dm3

  • 0.20 mol/dm3

  • 0.40 mol/dm3

  • 0.80 mol/dm3

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2a3 marks

A student investigated the rate of reaction between magnesium ribbon and hydrochloric acid.

Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

The experiment involved changing the concentration of the acid and observing its effect on the rate of reaction. 

The student did this by measuring the time taken for the magnesium to stop reacting. She set up the experiment as shown in Figure 1.

Figure 1

aqa-gcse-3-4m-q2a-magnesium-ribbon-changing-conc-exp

The concentration and results of the experiment were recorded in Table 1.

Table 1

 

Concentration / mol dm-3

Time for reaction to finish / s

Test tube 1

0.5

78

Test tube 2

1.0

62

Test tube 3

1.5

50

Test tube 4

2.0

36

State three variables that the student should control.

2b3 marks

The rate of reaction increased as the concentration of hydrochloric acid increased. 

Explain why increasing concentration has this effect.

2c2 marks

Suggest how the student prepared solutions with differing concentrations.

2d6 marks

The student had a stock solution of sodium hydroxide with an exact concentration of 0.200 moles per dm3. 

She wanted to use this solution to find the concentration of a solution of hydrochloric acid.

She used a pipette to transfer 10.00 cm3 of the hydrochloric acid into a conical flask and filled a burette with the 0.200 moles per dm3 sodium hydroxide solution. 

Describe how a titration could be used to obtain accurate results.

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3a1 mark

Convert the volumes in parts (a) - (e). 

1050 cm3 = _____ dm3

3b
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1 mark

865 cm3 = _____ dm3

3c
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1 mark

37250 cm3 = _____ dm3

3d
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1 mark

__ cm3 = 1.63 dm3

3e
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1 mark

_____ cm3 = 0.805 dm3

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4a2 marks

A student performs a series of titrations and gets the following values: 24.7 cm3, 26.1 cm3, 24.8 cm3. 

Which values are concordant and why?

4b3 marks

Why should the burette and pipette be washed out with their solutions before being used in the experiment?

4c3 marks

What is the purpose of performing a rough titration in the first place?

4d2 marks

Explain why Universal Indicator shouldn’t be used in a titration.

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5a3 marks

Calculate the concentration of the following solutions of NaOH in g dm-3. You must show your working.

Relative formula mass (Mr): NaOH = 40

1 mol NaOH in 500 cm3.

5b3 marks

2.5 mol NaOH in 2000 cm3

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6
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1 mark

How much solute is present in 2 dm3 of a 0.5 mol/dm3 solution of potassium iodide?

  • 0.5 mol

  • 1.0 mol

  • 2.0 mol

  • 10 mol

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7
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1 mark

In a titration 25.00 cm3 of 0.20 mol/dm3 sodium hydroxide reacted with 26.50 cm3 of hydrochloric acid.

HCl + NaOH  → NaCl + H2O

What is the concentration of the acid?

  • 0.19 mol/dm3

  • 0.23 mol/dm3

  • 0.55 mol/dm3

  • 0.82 mol/dm3

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8a3 marks

A chemist prepared a solution of ethanoic acid for titration.

The concentration of ethanoic acid in the solution was 0.60 moles per cubic decimetre.

Calculate the mass in grams of ethanoic acid (CH3COOH) in 250 cm3 of this solution.

The relative formula mass (Mr) of ethanoic acid = 60.

Mass of ethanoic acid = ___________________ g

8b4 marks
0.75 mol of NaOH is dissolved in 2 dm3 of water. 

 

Write down the formula for calculating concentration and use it to find the concentration of this solution. 

Include the correct units. 

Relative formula mass (Mr): NaOH = 40

Concentration = ___________________

8c3 marks

The chemist used a 0.150 mol / dm3 solution of NaOH for the titration with ethanoic acid.

Calculate the mass of sodium hydroxide in 45.0 cm3 of this solution. 

Relative formula mass (Mr): NaOH = 40

Mass of sodium hydroxide = ___________________ g

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1a
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3 marks

This question is about citric acid (C6H8O7).

Citric acid is a solid.

A student used a solution of sodium hydroxide to determine the concentration of a solution of citric acid by titration.

The student made a 250 cm3 standard solution of sodium hydroxide of concentration 0.150 mol / dm3

Calculate the mass of sodium hydroxide required.

Relative atomic masses (Ar):     Na = 23,     O = 16,     H = 1

1b3 marks

The student used this method to titrate citric acid with sodium hydroxide solution.

  1. Pipette 25.0 cm3 of citric acid solution into a conical flask.
  2. Add a few drops of thymol blue indicator to the citric acid solution.
    Thymol blue is blue in alkali and yellow in acid.
  3. Add sodium hydroxide solution from a burette until the end-point is reached.

Explain what would happen at the end-point of this titration.

1c2 marks

Give two reasons why the sodium hydroxide solution is measured using a burette.

1d
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5 marks

The student's titration results are shown in Table 1.

 Table 1

 

 Titration

1

2

3

4

5

Volume of sodium hydroxide solution in cm3

24.60

23.05

23.15

23.30

23.10

 

The equation for the reaction is:

 C6H8O7 + 3NaOH ⟶ C6H5O7Na3 + 3H2O

Concordant results are those within 0.10 cm 3 of each other.

Calculate the concentration of the citric acid solution in mol / dm 3

Use only the concordant results from the table in your calculation.

You must show your working.

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2a
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1 mark

A student attempted to make pure zinc chloride crystals using the following method.

  • Add 5.00 g of zinc carbonate to 25.0 cm3 hydrochloric acid of 1.25 mol / dm3 concentration
  • Filter the mixture
  • Leave until crystals form

Explain why the student must add the zinc carbonate in excess. 

 
2b
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3 marks

The student then completed three separate experiments to produce zinc chloride by reacting 15.0 g of zinc carbonate with hydrochloric acid. 

ZnCO3 (s) + 2HCl (aq) →  ZnCl2 (aq) + CO2 (g) + H2O (l)

The highest percentage yield of zinc chloride that the student achieved in a single experiment was 82.4%. 

Calculate the mass of zinc chloride the student actually produced in this experiment. 

Relative atomic masses (Ar):     C = 12,     Zn = 65,     O = 16,     Cl = 35.5,     H = 1

2c
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1 mark
The lowest yield of zinc chloride crystals that the student achieved was 10.86 g.

 

The student used these zinc chloride crystals to make a 250 cm3 solution.

 Calculate the concentration of this solution in g / dm3.

2d
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2 marks
The student's third experiment produced 12.24 g of zinc chloride.

 

The student used these zinc chloride crystals to make a 50.0 cm3 solution. 

Calculate the concentration of this solution in mol / dm3.

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3
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1 mark

A student titrated 25.00 cm3 of 0.40 mol/dm3 potassium hydroxide solution against sulfuric acid and found that 23.55 cm3 of the acid was required to reach the end point.

H2SO4 + 2KOH  →  K2SO4 + 2H2O

What is the concentration of the sulfuric acid?

  • 0.42 mol/dm3

  • 0.105 mol/dm3

  • 0.35 mol/dm3

  • 0.21 mol/dm3

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4a4 marks

This question is about the reactions of copper compounds.

A student was asked to make copper(II) oxide, CuO, from copper(II) sulfate, CuSO4 using the following method.

  • Measure out 25.0 cm3 of 0.500 mol / dm3 copper(II) sulfate solution
  • Add an excess of sodium hydroxide solution
  • Do not filter out the copper(II) hydroxide, Cu(OH)2, precipitate that forms
  • Carefully heat the reaction mixture to decompose the copper(II) hydroxide

Write two balanced symbol equations, including state symbols, for the two reactions that occur.

4b
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3 marks

Calculate the number of grams of copper(II) sulfate present in the student's solution. 

Give your answer to three significant figures. 

Relative atomic masses (Ar):     Cu = 63.5,     S = 32,     O = 16

4c
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3 marks

What mass of copper(II) oxide will the student produce? 

Give your answer to three significant figures.

4d2 marks

Another student produces 1.27 g of pure, dry copper(II) oxide. 

The student's copper(II) oxide is placed into a 100 cm3 conical flask. After adding 50.0 cm3 of distilled water, the conical flask is stirred. 

The student completed a calculation and concluded that they had made a 25.4 g / dm3 solution. 

Explain how the student achieved this concentration and why it is incorrect.

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5a
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5 marks

This question is about acids and alkalis.

A student titrated 25.0 cm3 portions of sodium hydroxide solution with a 0.121 mol / dm3 dilute sulfuric acid.

Table 1 shows the student’s results. 

Table 1
 

 

Titration
1

Titration
2

Titration
3

Titration
4

Titration
5

Volume of dilute sulfuric acid in cm3

11.75

11.10

11.05

10.55

11.10

 

The equation for the reaction is: 

2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O 

Calculate the concentration of the sodium hydroxide in mol / dm3  

Use only the student’s concordant results. 

Concordant results are those within 0.10 cm3 of each other.

5b3 marks

A second student plans to perform the same titration described in part (a). 

2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O 

The second student will replace the sodium hydroxide solution of unknown concentration with a potassium hydroxide solution that has the exact same unknown concentration.

The equation for the new reaction is: 

2KOH + H2SO4 ⟶ K2SO4 + 2H2O

Before the student starts, they state that the concentration of the potassium hydroxide solution will be the same as the concentration of the sodium hydroxide solution.

State the assumption that the student makes in their statement and explain whether the student is correct.

5c2 marks
A third student plans to perform the same titration described in part (a).

 

2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O 

They intend to replace the sulfuric acid with hydrochloric acid of the same 0.121 mol / dm3 concentration. 

The equation for the new reaction is: 

NaOH + HCl ⟶ NaCl + H2O 

The final concentration of the sodium hydroxide should be the same as part (a). 

Explain the difference that the student can expect to see in their experimental results.

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6
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1 mark

Sodium hydroxide solution is neutralised by citric acid.

The equation for the reaction is:

C6H8O7 + 3NaOH ⟶ C6H5O7Na3 + 3H2O

Calculate the mass of the citric acid needed to neutralise 12.0 cm3 of 1.02  mol/dm3 sodium hydroxide solution.

Relative formula mass (Mr):  C6H8O7 = 192

  • 0.783 g

  • 0. 852 g

  • 1.22 g

  • 2.35g

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7a2 marks

Sodium hydroxide neutralises sulfuric acid.

The equation for the reaction is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

Write the ionic equation, including state symbols, for this reaction.

7b2 marks

Explain why the reaction of sodium hydroxide with sulfuric acid is not a redox reaction.

7c
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2 marks
Sodium hydroxide solution is considered corrosive at concentrations above 0.5 mol / dm3 and a low hazard at concentrations below 0.125 mol / dm3.

Concentrations between these values should be labelled as a health hazard as they can be irritating to the skin and eyes. These concentrations are acceptable for use in the school laboratory, although the lowest possible concentration of sodium hydroxide solution should always be used.  

Calculate the maximum and minimum mass of sodium hydroxide required to make a 500 cm3 sodium hydroxide solution that can be used in the school laboratory.

Relative atomic masses (Ar):     Na = 23,     O = 16,     H = 1

7d
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6 marks
A student was given a bottle of sodium hydroxide solution of unknown concentration.

 

The student used a pipette to transfer 25.0 cm3 of this sodium hydroxide solution into a conical flask for titration with 0.150 mol / dm3 sulfuric acid. 

The results of the student's titrations are shown in Table 1.

 
Table 1
 
 

Rough

Titration 1

Titration 2

Titration 3

Titration 4

Volume of 0.150 mol / dm3  sulfuric acid in cm3

25.25

24.75

25.20

25.00

25.15

 

Concordant results are within 0.10 cm3 of each other.

Use the student’s concordant results to determine if the bottle of sodium hydroxide solution of unknown concentration should be labelled as:

  • Corrosive
  • Health hazard
  • Low hazard

Show your working.

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