Is Matter Around Us Pure? — Practice

Question 1

Classify each as an element, compound or mixture: (a) Sodium chloride, (b) Brass, (c) Hydrogen, (d) Sugar solution.

Accepted Answer

Step 1 — Apply the definitions.
  Element: one type of atom; Compound: two+ elements in fixed ratio chemically bonded; Mixture: physical combination, variable composition.

Step 2 — Classify each.
  (a) NaCl: two elements (Na, Cl) chemically combined → COMPOUND.
  (b) Brass: Cu + Zn alloy, variable composition → HOMOGENEOUS MIXTURE.
  (c) H₂: one type of atom → ELEMENT.
  (d) Sugar dissolved in water: variable composition → HOMOGENEOUS MIXTURE.

✦ Answer: (a) compound, (b) mixture, (c) element, (d) mixture.

Question 2

In 50 g of a salt solution containing 5 g of salt, identify the solute, solvent and the mass of solvent.

Accepted Answer

Step 1 — Larger component = solvent; smaller = solute.
  Salt: 5 g (solute). Solvent: water.

Step 2 — Mass of solvent = mass of solution − mass of solute.
  = 50 − 5 = 45 g.

✦ Answer: Solute = salt (5 g). Solvent = water (45 g). Mass of solvent = 45 g.

Question 3

Why is the Tyndall effect seen in fog but not in salt solution?

Accepted Answer

Step 1 — Tyndall effect = light scattering by particles.

Step 2 — Particle size comparison.
  Fog: colloid (water droplets, 1 nm–1 μm) → big enough to scatter light → beam is visible.
  Salt solution: true solution (particles < 1 nm) → too small to scatter visible light → beam invisible.

✦ Answer: Fog is a colloid with particles large enough to scatter light; salt solution has particles too small (true solution) to scatter visible light.

Question 4

Name the property that distinguishes a suspension from a colloid.

Accepted Answer

Step 1 — Compare suspension and colloid.
  Both are heterogeneous mixtures, but suspension's particles are larger.

Step 2 — Identify the giveaway property.
  Suspension particles SETTLE when undisturbed; colloid particles do NOT settle.
  Additionally, suspension particles are RETAINED by filter paper; colloid particles PASS through ordinary filter paper.

✦ Answer: Suspension particles settle (and are filtered out by ordinary filter paper); colloid particles do not settle (and pass through filter paper). Either property is acceptable for 1 mark.

Question 5

Name the separation technique you would use to separate: (a) cream from milk, (b) iron filings from sand, (c) salt from sea water, (d) the dyes in a black-ink sample.

Accepted Answer

Step 1 — Identify each mixture type and pick the method.
  (a) Cream from milk: colloid, small particles → CENTRIFUGATION.
  (b) Iron filings from sand: magnetic + non-magnetic → MAGNETIC SEPARATION.
  (c) Salt from sea water: solid dissolved in liquid → EVAPORATION.
  (d) Dyes in ink: pigments dissolved in water/alcohol → CHROMATOGRAPHY.

✦ Answer: (a) Centrifugation, (b) Magnetic separation, (c) Evaporation, (d) Chromatography.

Question 6

A solution contains 40 g of sugar dissolved in 360 g of water. Calculate the mass percentage of the solution.

Accepted Answer

Step 1 — Compute mass of solution.
  Mass of solution = mass of solute + mass of solvent = 40 + 360 = 400 g.

Step 2 — Apply the formula.
  Mass % = (mass of solute / mass of solution) × 100
        = (40 / 400) × 100
        = 10 %.

✦ Answer: 10 %.

Trap to avoid: NEVER use 360 g (mass of solvent) in the denominator. Always use 400 g (mass of solution).

Question 7

You're told a sugar solution is 15% sugar by mass and you need 60 g of sugar. How much water should you add?

Accepted Answer

Step 1 — Let the total mass of solution be M grams.
  Mass % = (60 / M) × 100 = 15.

Step 2 — Solve for M.
  60 × 100 = 15 × M
  M = 6000 / 15 = 400 g.

Step 3 — Mass of water.
  Water = M − sugar = 400 − 60 = 340 g.

✦ Answer: Add 340 g of water (the total solution becomes 400 g, of which 60 g is sugar = 15 %).

Verification: 60/400 = 0.15 = 15 %. ✓

Question 8

Classify each as solution, colloid or suspension AND justify with one observation: (a) Soap-water lather, (b) Sand stirred in water, (c) Sugar in water.

Accepted Answer

Step 1 — Apply the size + behaviour tests.

(a) Soap lather: small air bubbles in water; particles ~ 1 nm–1 μm; doesn't settle; scatters light → COLLOID (foam).
Justification: Tyndall effect visible if you shine a torch through it; particles don't settle but lather isn't transparent.

(b) Sand in water: visibly heterogeneous; particles > 1 μm; settles down within seconds; filter paper retains it → SUSPENSION.
Justification: It settles on standing — the test for a suspension.

(c) Sugar in water: clear, uniform; particles < 1 nm; doesn't settle; passes filter paper; no Tyndall effect → TRUE SOLUTION.
Justification: Beam of light passes through without being visible.

✦ Answer: (a) Colloid, (b) Suspension, (c) Solution. Justifications as above.

Question 9

Why is the Tyndall effect more visible through a forest canopy in the early morning than at noon? Justify with two reasons.

Accepted Answer

Step 1 — Recall: Tyndall effect requires (i) colloidal-sized particles in the air, and (ii) a beam of light (so scattering is visible against a darker background).

Step 2 — Conditions in a forest at different times.
  Morning: cool & humid → many water-droplet aerosols (mist, fog) — abundant colloidal particles. Sun is low, light enters at an angle, the scene is dark below the canopy → bright beams stand out clearly.
  Noon: warmer + drier → fewer aerosols; sun overhead, ambient light is bright → contrast is lost, beams are not visible against the bright background.

✦ Answer: (i) Morning air has more colloidal particles (mist, water droplets) → more scattering. (ii) Slanting morning sunlight cuts through a darker forest interior, making the scattered beam visible by contrast; midday's overhead bright light washes the beam out.

Question 10

Two miscible liquids A and B have boiling points 56 °C and 78 °C respectively. Which separation technique would you choose, and why?

Accepted Answer

Step 1 — Check boiling-point difference.
  78 − 56 = 22 °C. This is LESS than 25 °C.

Step 2 — Rule.
  • Difference > 25 °C → simple distillation works.
  • Difference < 25 °C → use FRACTIONAL DISTILLATION (column with glass beads gives many tiny distillation cycles).

Step 3 — Apply.
  Since the difference is 22 °C (< 25 °C), use fractional distillation.

✦ Answer: Use fractional distillation, because the boiling-point difference (22 °C) is too small for simple distillation to separate them effectively. The packed column provides repeated cycles of evaporation-condensation, achieving sharper separation.

Question 11

Why is fractional distillation, not simple distillation, used to separate gases of liquefied air?

Accepted Answer

Step 1 — Recall the components of air after liquefaction.
  Nitrogen (BP −196 °C), Oxygen (BP −183 °C), Argon (BP −186 °C).

Step 2 — Compare boiling-point differences.
  N₂ vs O₂: 13 °C apart. O₂ vs Ar: 3 °C apart. All < 25 °C.

Step 3 — Apply the rule.
  Such small differences are impossible to separate by simple distillation (which gives only one evaporation-condensation cycle). A fractionating column allows many repeated cycles, ultimately separating each gas.

✦ Answer: Because the boiling-point differences of the air-gases (N₂, O₂, Ar) are all less than 25 °C — much too close for simple distillation. Fractional distillation gives the multiple equilibrium stages needed for sharp separation.

Question 12

Iron and sulphur are mixed in a 7:4 mass ratio and (a) stirred together, (b) heated strongly. Compare the products with respect to colour, magnetism and the effect of dilute H₂SO₄.

Accepted Answer

Step 1 — Identify what happens in each case.
  (a) Stirring alone produces a MIXTURE of iron + sulphur powder (no chemical change).
  (b) Strong heating triggers a chemical reaction: Fe + S → FeS (iron sulphide, a COMPOUND).

Step 2 — Compare on three properties.

(i) Colour:
     Mixture: yellow sulphur visible mixed with grey iron filings — heterogeneous appearance.
     Compound (FeS): uniform black mass.

(ii) Magnetism:
     Mixture: iron filings still attract a magnet — separable physically.
     Compound: NOT magnetic; iron is chemically combined and has lost its individual magnetic property.

(iii) Reaction with dilute H₂SO₄:
     Mixture: produces odourless H₂ gas (from the iron) + leaves unreacted sulphur.
     Compound: produces H₂S gas (rotten-egg smell) — a clearly different chemical behaviour.

Step 3 — Conclusion.
  Stirring keeps the elements' individual properties → mixture.
  Heating combines them at the atomic level → compound with brand-new properties.

✦ Answer: (a) Mixture: yellow + grey, magnetic, odourless gas with H₂SO₄. (b) Compound (FeS): uniformly black, non-magnetic, gives smelly H₂S gas with H₂SO₄.

This is the classic experiment proving that 'compound ≠ mixture of constituent elements'.

Question 13

You're given a mixture of common salt (NaCl), ammonium chloride (NH₄Cl) and sand. Design a stepwise process to get all three components pure.

Accepted Answer

Step 1 — Identify each component's distinguishing property.
  • Ammonium chloride: SUBLIMES on heating (solid → gas, no liquid).
  • Salt (NaCl): SOLUBLE in water, doesn't sublime.
  • Sand: INSOLUBLE in water, doesn't sublime.

Step 2 — Design the sequence.
  Step (a) Sublimation: heat the dry mixture in a covered china dish. NH₄Cl sublimes and re-deposits on a cool inverted funnel above → recover NH₄Cl.
  Step (b) Add water to the residue (salt + sand). Salt dissolves; sand doesn't.
  Step (c) Filter through filter paper.
          • Filtrate: salt solution.
          • Residue: sand. Wash, dry → recover SAND.
  Step (d) Evaporate the filtrate to dryness OR crystallise to recover NaCl.

Step 3 — Recap.
  Sublimation → Dissolution → Filtration → Evaporation. Four physical methods, three pure components recovered.

✦ Answer: Sublimation removes NH₄Cl first. Dissolution and filtration separate sand (residue) from salt (in solution). Evaporation/crystallisation recovers the salt. All three are recovered as pure substances.

Question 14

300 g of a sugar solution at 60 °C contains 60 g of sugar. (a) Find its mass percentage. (b) If cooled to room temperature 18 g of sugar separates out (crystallises), find the mass percentage of the saturated solution at room temperature.

Accepted Answer

Step 1 (a) — Mass % at 60 °C.
  Mass % = (60 / 300) × 100 = 20 %.

Step 2 (b) — After cooling, 18 g of sugar separates as solid.
  Sugar still dissolved = 60 − 18 = 42 g.
  Mass of saturated solution = original mass − separated solid = 300 − 18 = 282 g.

Step 3 — Mass % of saturated solution.
  = (42 / 282) × 100
  ≈ 14.89 % (≈ 14.9 %).

✦ Answer: (a) 20 %, (b) ≈ 14.9 %.

Insight: solubility decreases with temperature for most solid solutes, so cooling forces some out as crystals — the basis of recrystallisation purification.

Question 15

In a petroleum refinery, why is crude oil fractionally distilled in a tall column rather than batch-distilled, and what does the column's height correspond to physically?

Accepted Answer

Step 1 — Recall what's in crude oil.
  A complex mixture of many hydrocarbons with boiling points ranging from < 40 °C (petrol, gasoline) to > 350 °C (lubricants, asphalt). Many fractions have BPs within a few °C of each other.

Step 2 — Why fractional, not batch.
  Each plate/tray in the column is effectively one equilibrium stage. The taller the column, the more stages → the finer the separation. Batch distillation gives only ONE stage, which is hopeless for closely-boiling hydrocarbons.

Step 3 — What the height physically represents.
  As vapour rises, it cools. Hotter (heavier) fractions condense lower in the column. Lighter (lower BP) fractions condense higher up. Each tray collects a different 'fraction':
  • Bottom: bitumen, lubricating oil.
  • Middle: diesel, kerosene.
  • Top: petrol, LPG, refinery gases.
  Column height ↔ number of equilibrium stages ↔ sharpness of separation.

✦ Answer: Fractional distillation in a tall column provides many equilibrium stages (each tray = one cycle of evaporation + condensation), which is essential when boiling-point differences are small. Height ↔ stages: the taller the column, the finer the separation between fractions.

Practice — Is Matter Around Us Pure?

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