AQA
Specification 7405 (A-level Chemistry)
Paper 1 7405/1 – Physical and Inorganic Chemistry
Paper 2 7405/2 – Organic and Physical Chemistry
Total marks Paper 1: 105 | Paper 2: 105
Assessment Written exam, 2 hours each paper
Status SAMPLE – for revision and preparation purposes only
.
PAPER 1 – Physical and Inorganic Chemistry (7405/1)
Question 1 | Atomic Structure & Mass Spectrometry 6 marks
(a) The mass spectrum of a sample of bromine shows two peaks at m/z =
79 and m/z = 81.
(i) State what is meant by the term relative atomic mass.
Answer Mark Additional Guidance
The weighted mean mass of an atom of 1 Must include 'weighted
an element mean'. Allow 'average'.
compared to 1/12 of the mass of one 1 Must reference carbon-
atom of carbon-12. 12 (not just carbon).
(ii) Use
the data to calculate the relative atomic mass of bromine. The
relative abundance of
79Br is 50.7%.
Answer Mark Additional Guidance
Ar = (79 × 50.7 + 81 × 49.3) / 100 1 Allow correct
substitution of any
consistent percentages.
= 79.99 80.0 (accept 79.9 – 80.1) 1 Award 2 marks for
correct answer without
working.
(b) Explain why the mass spectrum of a noble gas such as argon
shows only a molecular ion peak.
Answer Mark Additional Guidance
,Noble gases exist as individual atoms 1
(monatomic) / do not form bonds
so the only species that can be 1 Allow: no covalent bonds
ionised in the mass spectrometer is to break.
the atom itself / no fragmentation
possible.
, Question 2 | Amount of Substance 8 marks
(a) A 2.50 g sample of CaCO3 is added to excess dilute HCl. Calculate the
volume of CO2 produced at RTP. (Mr
3 −1
CaCO3 = 100.1; molar volume at RTP = 24.0 dm mol )
Answer Mark Additional Guidance
Moles CaCO3 = 2..1 = 0.02498 1 ECF throughout. Accept
mol 2.50/100 = 0.0250.
Moles CO2 = moles CaCO3 = 0.02498 1 Must show awareness of
mol (1:1 ratio) 1:1 stoichiometry.
Volume = 0.02498 × 24.0 = 0.600 dm3 1 Accept 0.599 – 0.600
(600 cm3) dm3.
(b) A solution contains 4.90 g of H SO per 250 cm3. Calculate its
concentration in mol
dm−3. (M H SO = 98.1) 2 4
r 2 4
Answer Mark Additional Guidance
Moles H2SO4 = 4..1 = 0.04995 mol 1
Concentration = 0..250 = 0.200 mol dm−3 1 ECF from moles; penalise if dm3 not
used.
(c) Define the term empirical formula.
Answer Mark Additional Guidance
The simplest whole-number ratio of 1 Do not accept
atoms of each element in a 'smallest'. Must state
compound. whole number ratio.
Question 3 | Bonding and Structure 9 marks
(a) Explain, in terms of structure and bonding, why graphite conducts
electricity but diamond does not.
Answer Mark Additional Guidance
Graphite: each carbon forms 3 1 Must mention
covalent bonds leaving one delocalised electrons.
delocalised electron per atom.
Delocalised electrons are free to move 1
through the layers and carry charge.
Diamond: each carbon forms 4 1 Accept: no mobile
covalent bonds; all electrons are charge carriers.
Specification 7405 (A-level Chemistry)
Paper 1 7405/1 – Physical and Inorganic Chemistry
Paper 2 7405/2 – Organic and Physical Chemistry
Total marks Paper 1: 105 | Paper 2: 105
Assessment Written exam, 2 hours each paper
Status SAMPLE – for revision and preparation purposes only
.
PAPER 1 – Physical and Inorganic Chemistry (7405/1)
Question 1 | Atomic Structure & Mass Spectrometry 6 marks
(a) The mass spectrum of a sample of bromine shows two peaks at m/z =
79 and m/z = 81.
(i) State what is meant by the term relative atomic mass.
Answer Mark Additional Guidance
The weighted mean mass of an atom of 1 Must include 'weighted
an element mean'. Allow 'average'.
compared to 1/12 of the mass of one 1 Must reference carbon-
atom of carbon-12. 12 (not just carbon).
(ii) Use
the data to calculate the relative atomic mass of bromine. The
relative abundance of
79Br is 50.7%.
Answer Mark Additional Guidance
Ar = (79 × 50.7 + 81 × 49.3) / 100 1 Allow correct
substitution of any
consistent percentages.
= 79.99 80.0 (accept 79.9 – 80.1) 1 Award 2 marks for
correct answer without
working.
(b) Explain why the mass spectrum of a noble gas such as argon
shows only a molecular ion peak.
Answer Mark Additional Guidance
,Noble gases exist as individual atoms 1
(monatomic) / do not form bonds
so the only species that can be 1 Allow: no covalent bonds
ionised in the mass spectrometer is to break.
the atom itself / no fragmentation
possible.
, Question 2 | Amount of Substance 8 marks
(a) A 2.50 g sample of CaCO3 is added to excess dilute HCl. Calculate the
volume of CO2 produced at RTP. (Mr
3 −1
CaCO3 = 100.1; molar volume at RTP = 24.0 dm mol )
Answer Mark Additional Guidance
Moles CaCO3 = 2..1 = 0.02498 1 ECF throughout. Accept
mol 2.50/100 = 0.0250.
Moles CO2 = moles CaCO3 = 0.02498 1 Must show awareness of
mol (1:1 ratio) 1:1 stoichiometry.
Volume = 0.02498 × 24.0 = 0.600 dm3 1 Accept 0.599 – 0.600
(600 cm3) dm3.
(b) A solution contains 4.90 g of H SO per 250 cm3. Calculate its
concentration in mol
dm−3. (M H SO = 98.1) 2 4
r 2 4
Answer Mark Additional Guidance
Moles H2SO4 = 4..1 = 0.04995 mol 1
Concentration = 0..250 = 0.200 mol dm−3 1 ECF from moles; penalise if dm3 not
used.
(c) Define the term empirical formula.
Answer Mark Additional Guidance
The simplest whole-number ratio of 1 Do not accept
atoms of each element in a 'smallest'. Must state
compound. whole number ratio.
Question 3 | Bonding and Structure 9 marks
(a) Explain, in terms of structure and bonding, why graphite conducts
electricity but diamond does not.
Answer Mark Additional Guidance
Graphite: each carbon forms 3 1 Must mention
covalent bonds leaving one delocalised electrons.
delocalised electron per atom.
Delocalised electrons are free to move 1
through the layers and carry charge.
Diamond: each carbon forms 4 1 Accept: no mobile
covalent bonds; all electrons are charge carriers.
