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Question 1

(a) Fig. 1.1 is a diagram of the female reproductive system in humans.
Identify the letter from Fig. 1.1 that represents the part where:
eggs are released 
fertilisation occurs 
implantation occurs 
meiosis occurs 
(b) A zygote divides to form an embryo.
Describe this type of cell division.
(c) During pregnancy the growing baby is supported by the placenta, umbilical cord, amniotic fluid and amniotic sac.
(i) State the function of the amniotic fluid.
(ii) Describe the function of the placenta and the umbilical cord.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic B15.4 — Sexual reproduction in humans (Part (a) — reproductive organs, fertilisation)
• Topic B16.2 — Cell division / mitosis (Part (b), (c)(i) & (c)(ii))

▶️ Answer/Explanation

(a) eggs released: B; fertilisation: A; implantation: E; meiosis: B

Eggs are released from the ovary, labelled B.
The egg travels up the oviduct (A), where sperm meet it and fertilisation occurs.
The fertilised egg (embryo) implants in the lining of the uterus wall (E).
Meiosis occurs in the ovary (B) during the production of egg cells.

(b) Mitosis

The zygote divides by mitosis, a type of nuclear division.
This produces genetically identical daughter cells.
It involves duplication of chromosomes before each division, so the number of chromosomes stays constant.

(c)(i) Protection

Amniotic fluid protects the fetus from mechanical damage/shock.

(c)(ii) Placenta and umbilical cord functions

The placenta acts as a barrier, preventing toxins and pathogens from crossing into fetal blood.
The umbilical cord carries oxygen and nutrients from the mother’s blood to the fetus.
It also carries carbon dioxide and excretory products from the fetus back to the mother.
Exchange of these substances across the placenta occurs by \( \text{diffusion} \).

Question 2

The element carbon exists as 3 naturally occurring isotopes.
Fig. 2.1 shows an atom of one isotope, carbon-14.
(a)(i) Complete the labels on Fig. 2.1.
(ii) Fig. 2.2 shows an atom of a different isotope of carbon.
Complete Fig. 2.2 to show the particles in the nucleus of one of the other two isotopes of carbon.
(iii) The different isotopes of carbon all have the same chemical properties. Explain why.
(b) Carbon reacts with oxygen to form carbon dioxide.
State the test for carbon dioxide and its positive result.
(c) Compounds that only contain carbon and hydrogen can form compounds with only single covalent bonds.
Complete the sentence about these compounds.
Choose words from the list.
addition        alkenes        hydrocarbons
polymers        saturated        unsaturated
Carbon and hydrogen compounds with only single covalent bonds are called ………………………… ………………………… .

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C2.2 — Atomic structure and the Periodic Table (Part (a)(i) — proton, neutron, electron labels)
• Topic C2.3 — Isotopes (Part (a)(ii) and (a)(iii))
• Topic C12.5 — Identification of ions and gases (Part (b) — test for CO\(_2\))
• Topic C11.4 — Alkanes (Part (c) — saturated hydrocarbons)

▶️ Answer/Explanation

(a)(i) Electron, proton, neutron

The outer particles orbiting the nucleus are electrons.
The positively charged particles in the nucleus are protons.
The uncharged particles in the nucleus are neutrons.

(a)(ii) 6 protons; 6 or 7 neutrons

The diagram shows 6 electrons, so the atom must have 6 protons (same atomic number, since it’s still carbon).
Since it is a different isotope of carbon-14, the neutron number must differ, so either 6 or 7 neutrons is accepted.

(a)(iii) Same number of electrons in the outer shell

Isotopes have the same number of protons and therefore the same number of electrons.
Since chemical properties depend on the electronic configuration (especially the outer shell), isotopes react identically.

(b) Test: limewater; Positive result: turns milky/cloudy

Carbon dioxide gas is bubbled through limewater (\( \text{Ca(OH)}_2 \)).
A white precipitate of calcium carbonate forms, turning the limewater milky/cloudy.

(c) Saturated hydrocarbons

Compounds containing only carbon and hydrogen are called hydrocarbons.
When all the bonds are single covalent bonds, the compound is described as saturated.

Question 3

Fig. 3.1 shows a circuit used by students investigating how the resistance of a metal wire varies with length.
(a) The students use an ammeter and a voltmeter to measure the current in and potential difference across the wire.
(i) Complete Fig. 3.1 to show the correct symbols and positions for the ammeter and the voltmeter.
(ii) When the wire is 20 cm long, the ammeter reads 0.80 A and the voltmeter reads 3.0 V. Calculate the resistance of the wire, stating the correct unit.
(iii) On Fig. 3.2 sketch a graph to show how the resistance of the wire varies with length.
(b) The student notices that when the circuit is left switched on, the wire becomes warm.
Describe how conduction transfers thermal energy in the metal wire.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P4.3.1 — Circuit diagrams and circuit components (Part (a)(i))
• Topic P4.2.4 — Resistance (Part (a)(ii) and (a)(iii))
• Topic P2.3.1 — Conduction (Part (b) — this is a thermal physics topic, not electricity)

▶️ Answer/Explanation

(a)(i) Ammeter in series, voltmeter in parallel across the wire

The ammeter must be placed in series in the main circuit loop, since current is the same everywhere in a series circuit.
The voltmeter must be connected in parallel across the wire, since it measures the potential difference across that component only.

(a)(ii) 3.8 Ω

Resistance is calculated using \( R = \dfrac{V}{I} \).
\( R = \dfrac{3.0}{0.80} = 3.8 \, \Omega \).

(a)(iii) Straight line through the origin with positive gradient

Resistance of a wire is directly proportional to its length.
So the sketch should be a straight line starting at the origin with a positive gradient.

(b) Conduction via vibrating atoms and free electrons

As current flows, the atoms in the metal lattice vibrate more.
These vibrations are passed from atom to atom along the wire.
Additionally, free (delocalised) electrons move through the metal, transferring thermal energy rapidly.

Question 4

(a) A student investigates the effect of humidity on the rate of water uptake in plant shoots.
Fig. 4.1 shows the apparatus they use.
The student measures the distance moved by the air bubble in 2 minutes.
The student then covers the plant shoot with a plastic bag to increase the humidity and repeats the investigation.
The results are used to calculate the rate of movement of the air bubble.
Table 4.1 shows their results.
(i) Calculate the rate of movement of the air bubble at low humidity and complete Table 4.1.
(ii) The rate of water uptake is approximately equivalent to the rate of transpiration.
Complete the sentences to describe and explain the results in Table 4.1.
When the humidity is increased the distance moved by the air bubble in 2 minutes decreased by ……………………………………. mm.
Higher humidity means the concentration of water vapour in the air around the leaf increases.
This decreases the concentration ……………………………………. between the inside and the outside of the leaf.
Less water is lost from the surfaces of the mesophyll cells by the process of ……………………………………. .
Less water vapour diffuses through the ……………………………………. .
(iii) Suggest why not all the water taken up by the roots is lost to the atmosphere.
(iv) State one other factor that affects the rate of transpiration.
(b) Xylem vessels can draw up a column of water through transpiration pull.
(i) Describe how transpiration pull causes the movement of water molecules.
(ii) State the term used to describe how water molecules are held together in the column of water.
(iii) State one other substance transported in the xylem.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B8.2 — Water uptake (Part (a)(i) — apparatus and rate calculation)
• Topic B8.3 — Transpiration (Part (a)(ii), (a)(iii), (a)(iv), and (b)(i))
• Topic B8.1 — Xylem and phloem (Part (b)(ii) — cohesion, and (b)(iii) — mineral ions)

▶️ Answer/Explanation

(a)(i) 10.5 mm/min

Rate = distance moved ÷ time taken.
Rate = \( \dfrac{21.0}{2} = 10.5 \) mm per minute.

(a)(ii) Decreased by 16.0 mm; water potential/concentration gradient; evaporation; stomata

The distance moved decreased from 21.0 mm to 5.0 mm, a decrease of 16.0 mm.
Higher humidity reduces the water vapour concentration gradient between the leaf’s air spaces and the outside air.
Less water evaporates from the mesophyll cell surfaces into the air spaces.
Less water vapour then diffuses out through the stomata.

(a)(iii) Some water is used by the plant

Some of the water is used in photosynthesis or to keep cells turgid, supporting the plant structurally.

(a)(iv) Temperature (or wind speed)

Increasing temperature or wind speed increases the rate of transpiration.

(b)(i) Water potential gradient pulls water up

As water evaporates from the leaf, it creates a water potential gradient between the top and bottom of the xylem.
This pulls the column of water molecules upward through the xylem.

(b)(ii) Cohesion

Water molecules are held together by hydrogen bonds, a property known as cohesion, which keeps the water column continuous.

(b)(iii) Mineral ions

Mineral ions (such as nitrate ions) are also transported in the xylem alongside water.

Question 5

Table 5.1 gives some information about the properties of the Group VII elements.
(a)(i) Predict the boiling point of chlorine. Write your answer in Table 5.1.
(ii) Predict the state at room temperature of bromine. Write your answer in Table 5.1.
(b) Bromine has a lower boiling point than iodine. Tick (✓) one box to show the correct explanation.
(c) Chlorine, \(\text{Cl}_2\), reacts with sodium bromide, NaBr, to form sodium chloride, NaCl, and bromine.
(i) Construct the balanced symbol equation for this reaction.
(ii) Sodium chloride is an ionic compound.
A sodium atom has electronic structure 2.8.1, and a chlorine atom has electronic structure 2.8.7.
Draw a dot-and-cross diagram to show the ions formed when sodium bonds with chlorine, including the charges on the ions.
(iii) Concentrated aqueous sodium chloride conducts electricity. Tick (✓) one box to show the correct explanation.
(iv) State the name of the product at the anode in the electrolysis of concentrated aqueous sodium chloride.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C8.3 — Group VII properties (Parts (a)(i), (a)(ii), and (b))
• Topic C6.2 — Rate of reactions (Part (c)(i))
• Topic C2.4 — Ions and ionic bonds (Part (c)(ii) — dot-and-cross diagram)
• Topic C4.1 — Electrolysis (Parts (c)(iii) and (c)(iv))

▶️ Answer/Explanation

(a)(i) Between −122 °C and −30 °C

Chlorine’s boiling point lies between fluorine’s (−188 °C) and bromine’s (59 °C), following the increasing trend down the group.
Any value in this inclusive range is accepted.

(a)(ii) Liquid

Bromine’s boiling point (59 °C) is above room temperature but its melting point is below it, so bromine exists as a liquid at room temperature.

(b) The forces between bromine molecules are weaker

Halogens are simple molecular substances held together by weak intermolecular forces.
These forces increase down the group as molecules get larger, so bromine (smaller molecule) has weaker forces than iodine, giving it a lower boiling point.

(c)(i) \( \text{Cl}_2 + 2\text{NaBr} \rightarrow 2\text{NaCl} + \text{Br}_2 \)

Chlorine displaces bromine from sodium bromide because chlorine is more reactive (higher up Group VII).
The equation must be balanced for both atoms and formula units.

(c)(ii) 

Sodium (2.8.1) loses one electron to form Na\(^+\) with structure 2.8.
Chlorine (2.8.7) gains that electron to form Cl\(^-\) with structure 2.8.8.
The diagram should show the transferred electron as a cross on the chloride ion, with correct + and − charges shown outside square brackets.

(c)(iii) Contains ions which can move

Concentrated aqueous sodium chloride conducts electricity because it contains mobile ions (Na\(^+\) and Cl\(^-\)) that can carry charge through the solution.

(c)(iv) Chlorine

At the anode (positive electrode), chloride ions lose electrons (oxidation) to form chlorine gas.

Question 6

Light is a transverse wave which is refracted by a transparent material.
(a) Fig. 6.1 shows the refraction of a ray of light as it enters a transparent block.
(i) The refractive index of the transparent block is 1.55.
The angle of incidence is 45°. Calculate the angle of refraction.
(ii) Information can be transmitted using total internal reflection of light in an optical fibre.
Fig. 6.2 shows a ray of light entering an optical fibre.
Complete the ray diagram on Fig. 6.2 to show how an optical fibre can transmit light along the fibre.
(iii) State what is meant by the term critical angle.
(b) Lasers are used to produce light of one single wavelength.
A battery powered laser has a power output of 0.0060 W and an efficiency of 40%.
(i) Calculate the power input provided by the laser’s batteries.
(ii) A battery of three 1.5 V cells in the laser provides 20.0 C of charge before the cells need replacing.
Calculate how long this battery will power the laser for.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P3.2.2 — Refraction of light (Parts (a)(i), (a)(ii), and (a)(iii) — refractive index, TIR in optical fibres, critical angle)
• Topic P1.6.3 — Energy resources / efficiency (Part (b)(i) — efficiency equation, note this sits under the Energy topic, not Electricity)
• Topic P4.2.2 — Electric current (Part (b)(ii) — \(Q = It\))

▶️ Answer/Explanation

(a)(i) 27°

Using \( n = \dfrac{\sin i}{\sin r} \), rearrange to get \( \sin r = \dfrac{\sin 45}{1.55} \).
\( r = \sin^{-1}(0.456) \approx 27^\circ \).

(a)(ii) 

The ray should be drawn bouncing off the internal walls of the fibre at equal angles each time, never escaping, until it exits at the far end.

(a)(iii) The angle of incidence at which the angle of refraction is 90°

The critical angle is the minimum angle of incidence needed for total internal reflection to occur (light travelling from a denser to a less dense medium).

(b)(i) 0.015 W

Efficiency = \( \dfrac{\text{power output}}{\text{power input}} \times 100 \).
\( 40 = \dfrac{0.0060}{\text{power input}} \times 100 \), so power input = \( \dfrac{0.0060}{0.40} = 0.015 \) W.

(b)(ii) 6000 s

Current \( I = \dfrac{P}{V} = \dfrac{0.015}{4.5} = 0.00333 \) A (using total e.m.f. of \(3 \times 1.5 = 4.5\) V).
Time \( t = \dfrac{Q}{I} = \dfrac{20.0}{0.00333} = 6000 \) s.

Question 7

(a) Fig. 7.1 is a diagram of a villus.
(i) State the name of the part labelled Y in Fig. 7.1.
(ii) Describe the function of the part labelled X in Fig. 7.1.
(iii) State where villi are found in the alimentary canal.
(b) Coeliac disease is a condition which causes the villi to become inflamed and flattened.
Explain why coeliac disease may cause weight loss.
(c) Table 7.1 shows some digestive enzymes, their substrates and product(s).
Complete Table 7.1.
(d) State two parts of the alimentary canal where protease is secreted.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):
• Topic B7.3 — Digestion (Part (c) — enzymes, substrates, products; and Part (d) — sites of protease secretion)

▶️ Answer/Explanation

(a)(i) Capillary (blood vessel)

The network of blood vessels running through the villus is a capillary network, which carries away absorbed nutrients.

(a)(ii) Absorption of fats

The part labelled X is the lacteal, which absorbs digested fats (fatty acids and glycerol) into the lymphatic system.

(a)(iii) Small intestine

Villi line the inner wall of the small intestine, increasing its surface area for absorption.

(b) Reduced absorption surface area causes malnutrition

Flattened villi mean a smaller surface area is available for absorption.
This leads to fewer nutrients being absorbed into the blood, resulting in weight loss.

(c) 

Amylase breaks down starch into simpler sugars.
Protease breaks down protein into amino acids.

(d) Stomach and small intestine (or pancreas)

Protease is secreted in the stomach (as pepsin) and in the small intestine (via pancreatic secretions), where protein digestion occurs.

Question 8

A student investigates the reaction between large marble chips and excess dilute hydrochloric acid.
Fig. 8.1 shows the apparatus they use.
The student measures the total volume of carbon dioxide gas every 30 seconds.
Fig. 8.2 shows a graph of the student’s results.
(a)(i) State the time at which the reaction stops.
(ii) The student repeats the experiment using 20 g of small marble chips instead of 20 g of large marble chips.
Sketch a line on Fig. 8.2 to show the results you would expect.
(b) The student repeats the experiment again.
This time the student uses:
  • the same mass of small marble chips
  • the same volume of hydrochloric acid
  • more concentrated hydrochloric acid.
Explain, using ideas about collisions between particles, why the reaction is faster.
(c) The reaction between marble chips and dilute hydrochloric acid is an example of an exothermic reaction.
Use the axes shown in Fig. 8.3 to draw and label the energy level diagram for this type of reaction.
Label:
  • the energy levels of reactants and products,
  • the energy change in the reaction
  • the activation energy of the reaction.
(d) When 5 g of marble chips, \(\text{CaCO}_3\), react with dilute hydrochloric acid, HCl, 2.2 g of carbon dioxide is produced.
\( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \)
Calculate the volume occupied by this 2.2 g of carbon dioxide gas.
The molar gas volume at room temperature and pressure is 24 dm\(^3\).
[\(A_r\) : C, 12; O, 16]

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C6.2 — Rate of reaction / collision theory (Parts (a)(i), (a)(ii), and (b))
• Topic C5.1 — Exothermic and endothermic reactions (Part (c) — energy level diagram)
• Topic C3.3 — The mole and the Avogadro constant (Part (d) — molar gas volume calculation)

▶️ Answer/Explanation

(a)(i) 240 s

The reaction stops when the volume of gas produced stops increasing (the curve levels off), which occurs at 240 s on the graph.

(a)(ii) Steeper line from the origin, levelling off at the same final volume (80 cm³)

Smaller marble chips have a larger total surface area, so the reaction is faster initially (steeper gradient).
Since the mass of marble chips and acid are unchanged, the final volume of gas produced remains 80 cm³.

(b) More particles per unit volume, more frequent collisions

A more concentrated acid has more acid particles per unit volume.
This increases the frequency of collisions between reacting particles.
More frequent collisions mean more successful collisions per second, speeding up the reaction.

(c) 

Since the reaction is exothermic, the products’ energy level must be drawn below the reactants’ energy level.
The energy change (\(\Delta H\)) is the vertical drop from reactants to products.
The activation energy (\(E_a\)) is the “hump” from reactants up to the peak of the curve.

(d) 1.2 dm³

\( M_r \) of \(\text{CO}_2 = 12 + (16 \times 2) = 44\).
Moles of \(\text{CO}_2 = \dfrac{2.2}{44} = 0.05\) mol.
Volume = \( 0.05 \times 24 = 1.2 \) dm³.

Question 9

Fig. 9.1 shows a simple d.c. motor with a coil of wire containing 100 turns.
(a) The current in the coil causes forces to act on the coil, which make it turn about its axis.
(i) Fig. 9.1 shows a force of 1.2 N acting at 90° to the coil, at a distance of 3.5 cm from the axis. Calculate the moment of the force on the coil.
(ii) Suggest how the magnitude of the force in (a)(i) changes when both the number of turns on the coil is doubled and the current is doubled.
(b) Fig. 9.2 shows a toy boat that uses a motor similar to Fig. 9.1 to propel it across a pond.
The toy boat has a mass of 0.60 kg and travels at a maximum speed of 3.0 m/s.
Calculate the maximum kinetic energy of the toy boat, stating the unit.
(c) Fig. 9.3 shows a speed–time graph for part of the toy boat’s journey.
(i) Use Fig. 9.3 to describe the motion of the toy boat for this part of the journey.
(ii) Suggest why the shape of this graph is not a realistic description of the motion of the toy boat at 1.5 minutes.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.5.2 — Turning effect of forces / moments (Part (a)(i))
• Topic P4.5.5 — The d.c. motor (Part (a)(ii) — effect of turns and current on force)
• Topic P1.6.1 — Energy / kinetic energy (Part (b))
• Topic P1.2 — Motion / speed-time graphs (Part (c)(i) and (c)(ii))

▶️ Answer/Explanation

(a)(i) 0.042 N m

Moment = force × perpendicular distance from pivot.
Convert 3.5 cm to 0.035 m.
Moment = \( 1.2 \times 0.035 = 0.042 \) N m.

(a)(ii) The force increases by a factor of 4

Doubling the number of turns doubles the force, and doubling the current also doubles the force.
Combined, the force increases by a factor of \(2 \times 2 = 4\).

(b) 2.7 J

Kinetic energy \( = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 0.60 \times 3.0^2 \).
\( = 2.7 \) J.

(c)(i) Constant acceleration then constant speed

For the first 1.5 minutes, the speed increases steadily, showing constant acceleration.
After 1.5 minutes, the speed stays at 3.0 m/s, showing constant (uniform) speed.

(c)(ii) The change would happen gradually, not instantly

In reality, a sudden change from acceleration to constant speed cannot happen instantaneously.
The graph should show a curve near 1.5 minutes rather than a sharp corner.

Question 10

(a) Table 10.1 shows the effect of adrenaline on blood glucose concentration.
(i) Calculate the percentage increase in blood glucose concentration after an injection of adrenaline.
(ii) Suggest the target organ of adrenaline that causes the change shown in Table 10.1.
(iii) State two other effects of adrenaline on the body.
(iv) State the name of the component of blood that transports adrenaline.
(b) Chemicals also control activities in plants.
(i) State one example of chemical control of plant growth in response to a stimulus.
(ii) State the name of the chemical that controls growth in plant shoots.
(iii) Complete the definition of the term growth.
Growth is a permanent increase in size and ………………… by an increase in ………………………… .

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic B13.2 — Hormones (Part (a)(i), (a)(ii), (a)(iii), and (a)(iv))
• Topic B1.1 — Characteristics of living organisms (Part (b)(iii) — definition of growth)

▶️ Answer/Explanation

(a)(i) 50%

Increase = \(1200 – 800 = 400\).
Percentage increase = \( \dfrac{400}{800} \times 100 = 50\% \).

(a)(ii) Liver

Adrenaline acts on the liver, causing it to convert stored glycogen into glucose, raising blood glucose concentration.

(a)(iii) Increased heart rate; widened pupils

Adrenaline prepares the body for “fight or flight” by increasing heart rate/pulse rate.
It also widens (dilates) the pupils to improve vision in a stressful situation.

(a)(iv) Plasma

Adrenaline, like other hormones, is transported dissolved in the blood plasma.

(b)(i) Phototropism (or gravitropism)

Phototropism is the growth response of a plant shoot toward light, controlled by a plant hormone.

(b)(ii) Auxin

Auxin is the hormone responsible for controlling growth in plant shoots.

(b)(iii) Permanent increase in size and dry mass, by an increase in cell size/number

Growth is defined as a permanent increase in size and dry mass.
This occurs through an increase in cell size and/or cell number.

Question 11

A scientist investigates food colourings using paper chromatography.
Fig. 11.1 shows the chromatogram produced.
The result for dye A is not shown.
(a) Identify which dyes, B, C or D, are in the food colouring X.
(b) The \(R_f\) value of a food colouring is calculated using the formula:
\( R_f = \dfrac{\text{distance travelled by substance}}{\text{distance travelled by solvent}} \)
Calculate the \(R_f\) value for dye B. Show your working.
(c) Food colouring A has an \(R_f\) value of 0.44. Calculate the distance travelled by food colouring A.
(d) The scientist makes a solution of food colouring B by dissolving 2.43 g of the food colouring in 200 cm\(^3\) of distilled water.
Calculate the concentration of the solution made in mol/dm\(^3\).
The relative molecular mass, \(M_r\), of the food colouring is 486.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic C12.3 — Chromatography (Parts (a), (b), and (c))
• Topic C3.3 — The mole and the Avogadro constant (Part (d) — concentration calculation)

▶️ Answer/Explanation

(a) B and C

The spots for X align at the same heights as the spots for dyes B and C, showing X is a mixture containing those two dyes.

(b) 0.90

\( R_f = \dfrac{\text{distance travelled by substance}}{\text{distance travelled by solvent}} = \dfrac{5.4}{6.0} = 0.90 \).

(c) 2.64 cm (or 2.6 cm)

Rearranging the \(R_f\) formula: distance = \(R_f \times\) distance travelled by solvent.
Distance = \(0.44 \times 6.0 = 2.64\) cm.

(d) 0.025 mol/dm³

Moles = \( \dfrac{\text{mass}}{M_r} = \dfrac{2.43}{486} = 0.005 \) mol.
Convert 200 cm³ to dm³: \(0.200\) dm³.
Concentration = \( \dfrac{0.005}{0.200} = 0.025 \) mol/dm³.

Question 12

Radon is a radioactive gas which occurs naturally in rocks and soil.
(a) Radon-222 \(({}^{222}_{86}\text{Ra})\) is an unstable isotope which decays by emitting an alpha particle.
(i) Use the correct nuclide notation to show the decay of radon-222:
\( {}^{222}_{86}\text{Ra} \rightarrow {}^{\phantom{0}}_{\phantom{0}}\text{Po} + {}^{\phantom{0}}_{\phantom{0}}\alpha \)
(ii) Draw lines to match an alpha particle with its correct characteristics. One line has been drawn as an example.
(iii) Complete Fig. 12.1 to show the path of an alpha particle as it travels through the electric field between two charged plates.
(b) A sample of radon gas is stored in a container with a fixed volume.
(i) Explain, in terms of molecular motion, why the pressure in the radon gas increases when the temperature is increased.
(ii) The volume of the container is 0.050 m\(^3\)
The density of the radon gas is 9.7 kg/m\(^3\).
Calculate the weight of the radon gas in the container.
The gravitational field strength, \(g\), is 10 N/kg.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P5.2.3 — Radioactive decay (Part (a)(i) — nuclide equation)
• Topic P5.2.2 — The three types of nuclear emission (Part (a)(ii))
• Topic P4.2.1 — Electrical charge (Part (a)(iii) — deflection in an electric field)
• Topic P2.1.3 — Pressure changes / kinetic particle model (Part (b)(i))
• Topic P1.4 — Density, combined with P1.3 — Mass and weight (Part (b)(ii))

▶️ Answer/Explanation

(a)(i) \( {}^{222}_{86}\text{Ra} \rightarrow {}^{218}_{84}\text{Po} + {}^{4}_{2}\alpha \)

Mass number balances: \(222 = 218 + 4\).
Atomic number balances: \(86 = 84 + 2\).

(a)(ii) 

An alpha particle consists of 2 protons and 2 neutrons, giving it a mass of 4 and a charge of +2.
Because it is large and highly charged, it has a high ionising ability but low penetrating power (stopped by paper).

(a)(iii) Curves toward the negative plate

Since the alpha particle is positively charged, it is attracted toward the negative plate and repelled by the positive plate.
The path should curve toward the negative (−) plate as it travels between them.

(b)(i) Increased particle speed causes more frequent, harder collisions

Raising the temperature increases the kinetic energy (speed) of the gas particles.
Faster particles collide with the container walls more often and with greater force.
This increases the force per unit area on the walls, i.e. the pressure.

(b)(ii) 4.9 N

Mass = density × volume = \( 9.7 \times 0.050 = 0.485 \) kg.
Weight = mass × g = \( 0.485 \times 10 = 4.9 \) N.

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