Chapter 5: Exploring Mixtures and their Separation (Chemistry)

Answer: (iv)

• Muddy water is a suspension: heterogeneous. Milk and blood are colloids: their particles are dispersed, not dissolved, so they are heterogeneous (even though they look uniform). Brass is an alloy (a solid solution of copper and zinc): homogeneous.
• (i) is wrong because smoke (a colloid of solid particles in air) is heterogeneous.
• (ii) is wrong: brass and vinegar are homogeneous, and muddy water is heterogeneous.
• (iii) is wrong because milk is heterogeneous.

(iv)

Answer: (iii) a and c

The Tyndall effect is the scattering of light by particles large enough to scatter it (colloids and suspensions).

• (a) Air and dust: dust particles are large and scatter light; a beam of sunlight entering a dark room becomes visible. Shows the effect.
• (c) Starch and water: forms a colloid; its particles (1–1000 nm) scatter light. Shows the effect.
• (b) Copper sulfate and water and (d) acetone and water are true solutions; their particles are smaller than 1 nm and cannot scatter light, so the beam's path is not visible.

So options (i), (ii) and (iv) are wrong because each contains (b) or (d).

(iii): a and c

Solution: tiny particles (< 1 nm), evenly distributed, do not settle, not filterable, transparent (salt solution, brass). Suspension: large particles (> 1000 nm) that settle, scatter light, filterable, heterogeneous (sand in water, mud). Colloid: 1–1000 nm, evenly distributed, do not settle, scatter light, not filterable, heterogeneous (milk, smoke, butter).

(i) All the components are solids measured by mass, so use mass by mass percentage (% m/m). Total mass $=75+420+5=500$ g.

$$\text{Sugar}=\frac{75}{500}\times 100=15\%,\text{Flour}=\frac{420}{500}\times 100=84\%,{\text{NaHCO}}_3=\frac{5}{500}\times 100=1\%$$

(ii) $$\text{Copper}=\frac{70}{100}\times 120=84\text{g},\text{Zinc}=120-84=36\text{g}$$

(i) % m/m: sugar 15%, flour 84%, sodium hydrogencarbonate 1%. (ii) Copper 84 g, zinc 36 g.

Density of the oil $=\frac{910\text{g}}{1000\text{mL}}=0.91$ g mL⁻¹, which is less than that of water (1 g mL⁻¹).

• Oil and water are immiscible, so they form two separate layers.
• The oil is lighter (less dense), so it forms the top layer; water stays at the bottom.
• They are separated with a separating funnel: pour the mixture into the funnel and let it stand until two clear layers form. Open the stopcock and let the lower water layer run into a beaker. Close the stopcock just as the oil reaches it; the oil is left in the funnel.

Yes, it forms a separate layer; oil (density 0.91 g mL⁻¹) floats on top of water. The two layers are separated with a separating funnel.

Answer: (iii)

A is true: the path of light is not visible in a solution. R is false: solution particles are very small (less than 1 nm), and that is exactly why they cannot scatter light. Particles of 100 nm would be colloidal and would scatter light.

(iii): A is true, R is false.

All the given mixtures can be separated, because in each the components differ in some physical property (size, density, solubility, volatility or rate of movement).

Mud: sedimentation, decantation and filtration. Plasma: centrifugation. Naphthalene and sand: sublimation. Chalk and salt: dissolve, filter, evaporate. Salt and water: evaporation. Oil and water: separating funnel. Flower pigments: paper chromatography.

The boiling points differ by $90-60=30$ °C, which is more than 25 °C, so they can be separated by simple distillation.

Heat the mixture in a distillation flask. When the thermometer reads about 60 °C, liquid A boils off; its vapour cools in the water condenser, and the liquid A collects in the receiver. The temperature stays near 60 °C until all of A has distilled; B (b.p. 90 °C) remains in the flask.

Simple distillation (boiling points differ by 30 °C); A distils first at about 60 °C and B remains in the flask.

Evaporation: solid needed, solvent not required. Crystallization: very pure solid needed (impurities stay in solution). Distillation: the liquid needs to be recovered, or miscible liquids with well-separated boiling points.

(i) The particles of a suspension settle down when left undisturbed. If blood were a true suspension, the blood cells would settle in the blood vessels wherever the flow is slow (or when we rest). The cells would not stay evenly spread, so oxygen and nutrients would not reach all parts evenly, and settled cells could clump and block vessels. Because blood is a colloid, its components stay uniformly dispersed.

(ii) Dispersed phase: blood cells (red blood cells, white blood cells) and platelets (along with proteins). Dispersion medium: plasma (mainly water).

(i) Blood cells would settle down, so blood would not flow and transport materials evenly and could block vessels. (ii) Dispersed phase: blood cells and platelets; dispersion medium: plasma.

In Fig. 5.25b, set-up 1 is sublimation (china dish heated with an inverted funnel over it), set-up 2 is evaporation (solution heated in a china dish) and set-up 3 is filtration (pouring into a funnel with filter paper).

1. Sublimation (1): heat the mixture; naphthalene sublimes and deposits on the cool inner wall of the funnel. Sand and salt remain.
2. Dissolving and filtration (3): add water to the remaining mixture and stir; salt dissolves. Filter: sand stays on the filter paper, salt solution passes through.
3. Evaporation (2): heat the filtrate; water evaporates and common salt is left.

1 → 3 → 2: sublimation (naphthalene), then dissolving in water and filtration (sand), then evaporation (salt).

Water and acetone are miscible (they form a solution), so they cannot be separated with a separating funnel or by filtration. Their boiling points are very different: acetone 56 °C and water 100 °C, a difference of 44 °C (more than 25 °C). On heating, acetone boils off first; its vapour is condensed and collected separately, while water remains in the flask.

Because they are miscible but have widely different boiling points (56 °C and 100 °C), so acetone distils over first, leaving water behind.

(i) At 40 °C, 62 g of potassium nitrate dissolves in 100 g of water. For 50 g of water:

$$\frac{62}{100}\times 50=31\text{g}$$

(ii) Crystals of potassium chloride separate out as the solution cools. Its solubility falls from 54 g per 100 g of water at 80 °C to about 36 g at 25 °C (between 35 g at 20 °C and 37.4 g at 30 °C). The water can no longer hold all the dissolved salt, so the excess, roughly 54 − 36 ≈ 18 g for every 100 g of water, crystallizes.

(iii) For these salts, solubility generally increases with temperature, but by very different amounts:

(i) 31 g (ii) KCl crystals appear, because solubility falls from 54 g to about 36 g per 100 g of water on cooling. (iii) Solubility increases with temperature: most for potassium nitrate, then ammonium chloride, then potassium chloride; sodium chloride hardly changes.

(i) Mass of solution = mass of sugar + mass of water.

$$A=\frac{20}{20+80}\times 100=20\%,B=\frac{20}{20+100}\times 100=16.67\%,C=\frac{30}{30+80}\times 100=27.27\%$$

(ii) Student C's solution is the most concentrated: it has the most solute (30 g) in the least solvent (80 g), so the largest mass of sugar per 100 g of solution.

(i) A = 20%, B = 16.67%, C = 27.27% (m/m) (ii) Student C.

(i) S = simple distillation.

(ii) A = distillation flask (round-bottom flask holding the mixture), B = condenser (water condenser), C = receiver (conical flask collecting the distillate).

(iii) Simple distillation needs a boiling point difference of at least about 25 °C (or a liquid with a dissolved solid):

(i) Simple distillation (ii) A: distillation flask, B: condenser, C: receiver (iii) (a) water–acetone and (b) water–salt.