AQA A-Level Chemistry Ultimate QUESTION Bank | 300+ Practice QUESTIONs, Model Answers & Detailed Explanations

AQA A-Level Chemistry Ultimate QUESTION Bank |
300+ Practice QUESTIONs, Model Answers &
Detailed Explanations

QUESTION 1
In a time of flight (TOF) mass spectrometer, a 1.20 × 10−21 g ion travels
through a flight tube of length 1.50 m with a kinetic energy of
2.40 × 10−15 J. What is the time of flight, in seconds?
• A. 2.37 × 10−5
• B. 1.18 × 10−5
• C. 4.74 × 10−5
• D. 3.50 × 10−5
Correct Answer: A. 2.37 × 10−5
Detailed Rationale: First, convert mass to kilograms: 𝑚 = 1.20 ×
1
10−24 kg. Using the kinetic energy formula 𝐾𝐸 = 𝑚𝑣 2 , rearrange to
2
2𝐾𝐸 2×2.40×10−15
find velocity 𝑣 = √ =√ ≈ 63,245.5 ms−1 . The time of
𝑚 1.20×10−24
𝑑 1.50
flight is 𝑡 = = ≈ 2.37 × 10−5 s.
𝑣 63,245.5

QUESTION 2

What is the volume, in dm3 , of 11.0 g of carbon dioxide gas measured
at 298 K and 100 kPa? (Ideal gas constant 𝑅 = 8.31 J K−1 mol−1 , 𝑀𝑟 of
CO2 = 44.0)

, • A. 3.10 dm3

• B. 6.19 dm3

• C. 6.24 dm3

• D. 12.4 dm3

Correct Answer: B. 6.19 dm3
11.0
Detailed Rationale: Calculate moles of CO2 : 𝑛 = = 0.25 mol. Using
44.0
𝑛𝑅𝑇
the ideal gas equation 𝑃𝑉 = 𝑛𝑅𝑇, rearrange for 𝑉: 𝑉 = =
𝑃
0.25×8.31×298
= 0.006186 m3 . Converting cubic meters to cubic
100,000
decimeters (1 m3 = 1000 dm3 ) gives 0.006186 × 1000 ≈ 6.19 dm3 .
QUESTION 3
Which of the following species has a tetrahedral shape?
• A. SF4
• B. XeF4
• C. PCl+
4

• D. H3 O+
Correct Answer: C. PCl+
4

Detailed Rationale: The PCl+4 ion has 4 bond pairs and 0 lone pairs
around the central phosphorus atom (total valence electrons = 5 + 4 −
1 = 8, divided by 2 = 4 pairs). According to VSEPR theory, 4 electron
pairs with no lone pairs form a regular tetrahedral shape.
QUESTION 4

,Given the standard enthalpy changes of formation (Δ𝑓 𝐻 ∘ ): CH4 (𝑔) =
−75 kJ mol−1 , CO2 (𝑔) = −393 kJ mol−1 , H2 O(𝑙) = −286 kJ mol−1 .
What is the standard enthalpy change of combustion for methane?

• A. −890 kJ mol−1

• B. −815 kJ mol−1

• C. +890 kJ mol−1

• D. −740 kJ mol−1

Correct Answer: A. −890 kJ mol−1
Detailed Rationale: The combustion equation is CH4 (𝑔) + 2O2 (𝑔) →
CO2 (𝑔) + 2H2 O(𝑙). Δ𝑐 𝐻 ∘ = ∑ Δ𝑓 𝐻 ∘ (products) −
∑ Δ𝑓 𝐻 ∘ (reactants) = [−393 + 2(−286)] − [−75 + 0] = (−393 −
572) − (−75) = −965 + 75 = −890 kJ mol−1 .
QUESTION 5
Which of the following statements about a Maxwell-Boltzmann
distribution curve for a gas at a constant temperature is correct?
• A. The area under the curve represents the total number of
molecules with energy greater than the activation energy.
• B. The curve starts at the origin because no molecules have zero
energy.
• C. The peak of radiation corresponds to the most probable energy
of the molecules.
• D. Raising the temperature shifts the peak of the curve to the right
and lower.

, Correct Answer: D. Raising the temperature shifts the peak of the curve
to the right and lower.
Detailed Rationale: When the temperature of a gas is increased,
molecules gain kinetic energy, causing the peak of the Maxwell-
Boltzmann distribution to shift to the right (higher energy) and
downwards (spread out over a broader range). The total area under the
curve represents the total number of molecules, not just those
exceeding activation energy.
QUESTION 6
For the reversible reaction 𝑁2 (𝑔) + 3𝐻2 (𝑔) ⇌ 2𝑁𝐻3 (𝑔), what are the
units of the equilibrium constant 𝐾𝑐 ?

• A. mol−2 dm6

• B. mol2 dm−6

• C. mol−1 dm3
• D. no units

Correct Answer: A. mol−2 dm6
[NH3 ]2
Detailed Rationale: The expression for 𝐾𝑐 is [N 3 . Substituting units:
2 ][H2 ]
2
(mol dm−3 ) 1
−3 3
= −3 2
= mol−2 dm6 .
(mol dm−3 )(mol dm ) (mol dm )

QUESTION 7
What is the oxidation state of chlorine in HClO3 ?
• A. +1
• B. +3