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CIE iGCSE Co-Ordinated Science C12.3 Chromatography Exam Style Questions Paper 4

CIE iGCSE Co-Ordinated Science Paper 4 - All Chapters

Question

A student investigates black ink using paper chromatography.
Fig. 5.1 shows:
  • the chromatogram the student obtains
  • the measurements the student may make.
(a) State if black ink is a pure or impure substance. Use Fig. 5.1 to explain your answer.
(b) State which two measurements on Fig. 5.1 are needed to calculate the \(R_f\) value of spot X.
(c) The student calculates the \(R_f\) value of spot Y to be 0.80.
The distance travelled by spot Y is 2.8 cm.
Calculate the distance travelled by the solvent.
(d) Paper chromatography has a stationary phase and a mobile phase.
The stationary phase is a solid.
The mobile phase is a liquid.
Describe what happens to the separation and motion of the particles when a solid changes to a liquid.
(e) Different substances have different structures.
Draw one line from each statement to the structure.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic C12.3 — Chromatography (Parts (a), (b) & (c))
• Topic C1.1 — Solids, liquids and gases (Part (d))
• Topic C2.5 — Simple molecules and covalent bonds / C2.6 — Giant covalent structures (Part (e))

▶️ Answer/Explanation

(a) Statement: Black ink is an impure substance.
Explanation: The chromatogram shows more than one spot, indicating that the ink contains multiple different substances (components).

(b) Measurements A and B.
\(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}\).
A is the distance travelled by spot X from the origin, and B is the distance travelled by the solvent front from the origin.

(c) Calculation:
\(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}\)
\(0.80 = \frac{2.8}{\text{distance travelled by solvent}}\)
Distance travelled by solvent = \(\frac{2.8}{0.80} = 3.5\text{ cm}\)

(d) Separation: Particles in a liquid are slightly further apart than in a solid (but still touching).
Motion: Particles in a liquid move faster than in a solid. They move randomly in a liquid, whereas they vibrate about a fixed position in a solid.

(e) Correct matching:

Question

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.

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