Pearson Edexcel International GCSE (9–1) · CHEMISTRY · Chapter
Chemical Formulae, Equations and Calculations Part 2
144 practice questions
Topics covered
- Calculations involving gas volumes19 practice questions
- Calculating the volume of a certain number of moles19 practice questions
- Using the molar volume in calculations with chemical equations19 practice questions
- Concentrations of solutions19 practice questions
- Calculations with equations involving solutions19 practice questions
- The standard calculation19 practice questions
- Reversing the calculations20 practice questions
Practice questions
21–30 of 61 · Normal
Question 21
4 marks · Normal
Answer and see the solution →The table below shows the volumes, in cm³, of a gas collected from several reactions at room temperature and pressure. Fill in the number of moles of gas for each row, using the molar volume of 24,000 cm³ at rtp.
Volume (cm³) Moles 6000 12,000 1800 24,000 Question 22
1 mark · Normal
Answer and check →Which calculation gives the number of moles of a gas from its volume at rtp (using cm³)?
- A.moles = volume / 24,000
- B.moles = mass / Mr
- C.moles = volume × 24,000
- D.moles = Mr / mass
Question 23
2 marks · Normal
Answer and see the solution →Calculate the volume, in dm³, of chlorine gas produced at rtp from the complete decomposition of 2.7 g of potassium chlorate, KClO₃. The balanced equation is:
(Ar: K = 39, Cl = 35.5, O = 16; molar volume = 24 dm³)
Question 24
4 marks · Normal
Answer and see the solution →Ammonium nitrate decomposes to form dinitrogen oxide (N2O) and water. The balanced equation is:
A practical uses 8.0 g of ammonium nitrate. (Ar: N = 14, H = 1, O = 16; molar volume at rtp = 24 dm³)
(a)[2 marks]Calculate the number of moles of ammonium nitrate used.
(b)[2 marks]Calculate the volume, in dm³, of dinitrogen oxide (N₂O) formed at rtp.
Question 25
2 marks · Normal
Answer and see the solution →A solution contains 4.0 g of potassium chloride (KCl) dissolved in exactly 100 cm³ of water. Calculate the concentration of this solution in g/dm³.
Question 26
4 marks · Normal
Answer and see the solution →(a)[1 mark]Give the formula that links concentration in mol/dm³, amount in mol, and volume in dm³.
(b)[1 mark]A student dissolves 2.5 g of magnesium sulfate (MgSO₄, Ar: Mg = 24, S = 32, O = 16) in 200 cm³ of water.
i. Calculate the number of moles of MgSO₄ in the solution.
(c)[2 marks]ii. Calculate the concentration of MgSO₄ in mol/dm³ in the solution.
Question 27
3 marks · Normal
Answer and see the solution →Calculate the mass of sodium nitrate (NaNO₃, Ar: Na = 23, N = 14, O = 16) required to make 250 cm³ of a 0.050 mol/dm³ solution.
Question 28
3 marks · Normal
Answer and see the solution →A teacher claims that doubling the amount of solute added to a fixed volume of water will always double the concentration of the solution. Explain using calculations and an example whether this statement is correct.
Question 29
4 marks · Normal
Answer and see the solution →Concentration and volume data for sodium sulfate solutions Sample Concentration (mol/dm³) Volume used (dm³) A 0.10 0.020 B 0.20 0.015 C 0.40 0.025 (a)[2 marks]Which sample contains the largest amount of sodium sulfate? Show your calculations.
(b)[1 mark]State the amount in mol present in that sample.
(c)[1 mark]Suggest one reason why a higher concentration solution may have less sodium sulfate than a lower concentration solution.
Question 30
1 mark · Normal
Answer and check →Which calculation gives the concentration, in mol/dm³, of a solution containing 1.5 mol of solute in 0.75 dm³ of solution?
- A.1.5 ÷ 0.75
- B.0.75 ÷ 1.5
- C.1.5 × 0.75
- D.0.75 × 1.5