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IGCSE · Cambridge (CIE) · 0970

Explain net diffusion from random particle movement

Separate individual random movements from the overall movement down a concentration gradient.

Diffusion pathway · 0970 (9–1) · 3.1
Exam scope and assessment notes

Core and Extended · Biology 0970, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0970 has the same subject content as IGCSE 0610, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.

Before you begin

  • Recall concentration as amount per unit volume.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What does a diffusion arrow normally show?

What you will learn

  • Separate individual random movements from the overall movement down a concentration gradient.
  • Apply the idea to a supplied biological example or investigation.
Equal-volume model: 18 particles cross A to B while seven cross back, giving net transfer of 11 toward B. Arrows summarise opposing transfers; they are not every individual particle trajectory.Random paths, net transferA: higherB: lower18 cross A → B7 cross B → ANet: 18 − 7 = 11Toward B · equal volumesRandom movement continues
Equal-volume model: 18 particles cross A to B while seven cross back, giving net transfer of 11 toward B. Arrows summarise opposing transfers; they are not every individual particle trajectory.

Count movement in both directions

Diffusion is the net movement of particles from a region where that substance has a higher concentration to one where it has a lower concentration. A concentration gradient is the difference in concentration across a distance. Molecules and ions move randomly; they do not sense a target or deliberately seek empty space.

Particles cross an imaginary boundary in both directions. When more of the same substance starts on one side, more tends to cross from that side in a given time. The difference between the two opposing movements is the net movement. Concentration means amount per unit volume, so compare equal volumes or given concentrations rather than raw counts in unequal containers.

Understand equilibrium

As the concentration difference diminishes, net diffusion decreases. At equal concentrations, random motion continues and particles still cross both ways, but there is no sustained net movement in this ideal model. Equal concentration is not the same as every particle becoming stationary.

A diagram arrow normally represents net transfer, not every individual path. Diffusion can occur without a membrane; a cell membrane is one possible boundary when the substance can pass through it.

Pause and explain

At equal concentrations, what happens to particles?

Put the idea to work

Worked example

In equal volumes A and B, a tracer has higher concentration in A. During one interval 18 tracer particles cross A to B and 7 cross B to A. Find the net transfer.

Show the worked solution
  1. Subtract the opposing transfers.
  2. 18 − 7 = 11 particles.
  3. The net direction is A to B, down the initial gradient.

Answer 11 particles net from A to B.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Over an interval 24 particles cross left to right and 9 cross back. Find net transfer.

Give me a hint

Subtract the reverse movement.

Compare my reasoning
  1. 24 − 9 = 15.
  2. The larger transfer is left to right.

15 particles net left to right.

Look for these in your work

  • Give magnitude and direction.
2 · Independent

At equilibrium 12 particles cross each way in an interval. State the net transfer and whether particles moved.

Give me a hint

Distinguish zero net transfer from zero movement.

Compare my reasoning
  1. 12 − 12 = 0 net.
  2. Both opposing crossings occurred.

Zero net transfer; particles still moved both ways.

Look for these in your work

  • State both parts.
3 · Transfer

A large compartment contains 100 particles in 10 cm³ and a small one contains 30 in 1 cm³. Predict initial net diffusion if they connect.

Give me a hint

Compare particles per cm³.

Compare my reasoning
  1. Large: 100 ÷ 10 = 10 per cm³.
  2. Small: 30 ÷ 1 = 30 per cm³.
  3. Net movement is from 30 toward 10.

From the small compartment to the large one.

Look for these in your work

  • Compare concentrations, not totals.

Common mistakes

  • Random movement occurs in both directions.
  • Use concentration, not just particle count in unequal volumes.
Recall without your notes

What can you explain now?

At equal concentrations, what happens to particles?

Compare with the explanation

They keep moving randomly, with no sustained net transfer.

Opposing transfers balance on average.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Explain why zero net diffusion does not mean zero particle motion.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Biology (9-1)

Course scope and assessment

780 sequenced lessons provide 2340 guided, independent and application problems in characteristics, classification, cells, organisation, specimen size, membrane transport, biological molecules, enzymes, plant and human nutrition, plant transport, circulatory systems, the heart, blood vessels, blood, disease transmission, body defences, immunity, gas exchange in humans, respiration energy uses and investigations, aerobic respiration, anaerobic respiration and recovery, excretion in humans, coordination and response, sense organs, hormones, homeostasis, tropic responses, drugs, asexual reproduction, sexual reproduction, sexual reproduction in plants, sexual reproduction in humans, sex hormones in humans, sexually transmitted infections, chromosomes, genes and proteins, mitosis, meiosis, monohybrid inheritance, variation, adaptive features, selection, energy flow, food chains and webs, nutrient cycles, population, community, ecosystem, population-growth phases, food supply, habitat destruction, pollution, conservation, biotechnology and genetic modification. 61 of 61 syllabus sections have linked teaching and staged practice. The earlier inheritance overview remains introductory, and the monohybrid pathway develops the relevant Core and Supplement requirements. This is not full programme completion or guaranteed exam readiness. Written practice is self-checked. This is the separate 9–1 qualification 0970; Cambridge states that its 2026–2028 subject content is otherwise the same as 0610. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. The 2029 syllabus is separately published at https://www.cambridgeinternational.org/Images/764352-2029-syllabus.pdf and is not yet mapped here. This course supports exams in 2026, 2027 and 2028 only.

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