Chapter 5: Exploring Mixtures and their Separation

Given data

• Mass of solute (salt), $$m_\text{solute} = 10\,\text{g}$$
• Mass of solvent (water), $$m_\text{solvent} = 90\,\text{g}$$

Step 1: Find the total mass of the solution.

By definition,

$$m_\text{solution} = m_\text{solute} + m_\text{solvent}$$

Substituting the given masses:

$$m_\text{solution} = 10\,\text{g} + 90\,\text{g} = 100\,\text{g}$$

Step 2: Write the formula for mass by mass percentage.

For a solution, the mass by mass percentage of solute is

$$\%\,m{\small /}m = \frac{m_\text{solute}}{m_\text{solution}} \times 100\%$$

Step 3: Substitute the values and simplify.

$$\%\,m{\small /}m = \frac{10\,\text{g}}{100\,\text{g}} \times 100\%$$

$$\%\,m{\small /}m = \frac{10}{100} \times 100\%$$

$$\%\,m{\small /}m = 10\%$$

Result

The solution therefore contains 10 percent salt by mass.

Step 1. Write the given data.

• Mass of solute (glucose), $$m_s = 5\,\text{g}$$
• Volume of the final solution, $$V = 100\,\text{mL}$$

Step 2. Recall the definition of mass by volume percentage.

The mass by volume percentage (also written as % (w/v)) tells us how many grams of solute are present in every $$100\,\text{mL}$$ of the solution. Its general formula is

$$\text{Mass by volume \%} = \left(\dfrac{\text{mass of solute (in g)}}{\text{volume of solution (in mL)}}\right) \times 100\%$$

Step 3. Substitute the given values.

$$\text{Mass by volume \%} = \left(\dfrac{5\,\text{g}}{100\,\text{mL}}\right) \times 100\%$$

Step 4. Do the arithmetic.

First divide:

$$\dfrac{5\,\text{g}}{100\,\text{mL}} = 0.05\,\text{g mL}^{-1}$$

Then multiply by $$100\%$$:

$$0.05 \times 100\% = 5\%$$

Step 5. Report the concentration.

The concentration of the glucose solution is therefore

Step 1 – Identify the known data

• Volume of the pesticide (solute): $$V_{\text{pest}} = 1\,\mathrm{mL}$$
• Total (final) volume of the spray (solution): $$V_{\text{soln}} = 100\,\mathrm{mL}$$

Step 2 – Recall the definition

The volume by volume percentage ( v/v %) of a component in a liquid mixture is defined as

$$\text{(v/v)\,\%} = \frac{\text{Volume of solute}}{\text{Total volume of solution}} \times 100$$

Step 3 – Substitute the numbers

$$\text{(v/v)\,\%} = \frac{1\,\mathrm{mL}}{100\,\mathrm{mL}} \times 100$$

Step 4 – Do the arithmetic

$$\text{(v/v)\,\%} = \frac{1}{100} \times 100 = 1$$

Step 5 – State the result

\[\boxed{\text{Volume by volume percentage of pesticide} = 1\,\%}\]

Given data

• Total mass of the talcum powder, $$m_{\text{powder}} = 300\,\text{g}$$.
• Mass percentage of zinc oxide (\( \mathrm{ZnO} \)) in the powder, $$w\% = 4\,\%\, (m/m)$$.

Step 1  — Translate the percentage into a fraction

By definition, a mass percentage relates the mass of a component to the total mass of the mixture:

$$\frac{\text{mass of } \mathrm{ZnO}}{\text{total mass of powder}} \times 100 = 4$$

This can be written in fractional form as

$$\frac{\text{mass of } \mathrm{ZnO}}{\text{total mass of powder}} = \frac{4}{100} = 0.04$$

Step 2  — Set up the proportion for 300 g of powder

Let $$m_{\mathrm{ZnO}}$$ be the mass of zinc oxide in 300 g of powder. Substitute the known mass of the powder:

$$\frac{m_{\mathrm{ZnO}}}{300}\times 100 = 4$$

Step 3  — Solve for $$m_{\mathrm{ZnO}}$$

First isolate $$m_{\mathrm{ZnO}}$$ by multiplying both sides by $$\frac{300}{100}$$:

$$m_{\mathrm{ZnO}} = \frac{4}{100}\times 300$$

Calculate the product:

$$m_{\mathrm{ZnO}} = 0.04\times 300 = 12$$

Step 4  — State the result

The mass of zinc oxide present in 300 g of the talcum powder is

\[ m_{\mathrm{ZnO}} = 12\,\text{g} \]

Step 1: Identify the data given.

• Each tablespoon of concentrate → $$15\,\text{mL}$$
• Number of tablespoons used → $$2$$
• Total volume of finished drink (juice) → $$150\,\text{mL}$$

Step 2: Calculate the volume of orange-juice concentrate added.

$$\text{Volume of concentrate}=2\times15\,\text{mL}=30\,\text{mL}$$

Step 3: Recall the formula for percentage by volume (%% $$v/v$$).

$$\%\,v/v=\left(\dfrac{\text{volume of solute (concentrate)}}{\text{total volume of solution}}\right)\times100$$

Step 4: Substitute the known values.

$$\%\,v/v=\left(\dfrac{30\,\text{mL}}{150\,\text{mL}}\right)\times100$$

Step 5: Do the arithmetic.

$$\dfrac{30}{150}=0.20$$

$$0.20\times100=20$$

Step 6: State the final result.

\[\boxed{\%\,v/v = 20\%}\]

Therefore, the orange-juice concentrate makes up 20 percent of the total volume of the prepared juice.

Given data

• Commercial vinegar must contain 5 % v/v acetic acid.
• Glacial acetic acid is 100 % (pure) acetic acid.

Step 1 – Recall the definition of volume %

For a solution expressed as "x % v/v"

$$\%\,v/v = \frac{\text{volume of solute (mL)}}{\text{total volume of solution (mL)}} \times 100$$

Step 2 – Set up the calculation

Let us prepare 100 mL of vinegar (any convenient volume can be chosen and then scaled).

Insert the known quantities:

$$5 = \frac{V_{\text{acetic acid}}}{100\,\text{mL}} \times 100$$

Step 3 – Solve for the required volume of glacial acetic acid

$$V_{\text{acetic acid}} = \frac{5 \times 100\,\text{mL}}{100} = 5\,\text{mL}$$

Step 4 – Determine the volume of water to add

Total desired volume = 100 mL

Volume already occupied by acetic acid = 5 mL

Therefore, volume of water required:

$$V_{\text{water}} = 100\,\text{mL} - 5\,\text{mL} = 95\,\text{mL}$$

Step 5 – Laboratory procedure

1. Put on gloves and goggles.
2. Measure 95 mL of distilled water in a 100 mL measuring cylinder or volumetric flask.
3. Slowly add 5 mL of glacial acetic acid into the water (never the reverse: "add acid to water" for safety).
4. Mix thoroughly. The final volume is 100 mL, giving 5 % v/v vinegar.

Optional scaling

Conclusion

To convert glacial acetic acid into table vinegar, dilute it so that 1 part acid is present in 19 parts water, giving a final concentration of 5 % v/v.

Step 1 : Read the solubility data from Activity 5.2
From the graph the approximate solubilities (g of solute per 100 g of water) are

• Compound A : $$S_{A,80}=140$$ at 80 °C and $$S_{A,60}=104$$ at 60 °C
• Compound B : $$S_{B,80}=38$$ at 80 °C and $$S_{B,60}=37$$ at 60 °C

Step 2 : Choose an equal mass of each hot, saturated solution
Take 100 g of each saturated solution at 80 °C (the problem says “equal masses”).
For any solute with solubility $$S$$, the composition of 100 g of saturated solution is :

• Mass of water : $$m_w = \frac{100}{100+S}\times100\;\text{g}$$
• Mass of solute : $$m_{s,80}= \frac{S}{100+S}\times100\;\text{g}$$

Step 3 : Calculate for compound A

• At 80 °C : $$m_{s,80}(A)= \frac{140}{100+140}\times100 = 58.3\,\text{g}$$ and $$m_w(A)=41.7\,\text{g}$$
• At 60 °C the same 41.7 g of water can now hold at most
  $$m_{s,60\,\max}(A)= \frac{104}{100}\times41.7 = 43.3\,\text{g}$$
• Solid deposited : $$m_p(A)=58.3-43.3 = 15\,\text{g}$$

Step 4 : Calculate for compound B

• At 80 °C : $$m_{s,80}(B)= \frac{38}{100+38}\times100 = 27.5\,\text{g}$$ and $$m_w(B)=72.5\,\text{g}$$
• At 60 °C : $$m_{s,60\,\max}(B)= \frac{37}{100}\times72.5 = 26.8\,\text{g}$$
• Solid deposited : $$m_p(B)=27.5-26.8 = 0.7\,\text{g}$$

Step 5 : Compare
$$m_p(A)=15\,\text{g} \gg m_p(B)=0.7\,\text{g}$$
Hence the saturated solution of compound A deposits far more solid on cooling.

Concept recalled – Crystallisation

When a hot, saturated solution of common salt (sodium chloride, $$\mathrm{NaCl}$$) is allowed to cool and the solvent (water) subsequently evaporates, the excess $$\mathrm{NaCl}$$ separates out as solid crystals. Two simultaneous processes decide the size of the final crystals:

• Nucleation: first tiny particles (nuclei) of the solid appear.
• Crystal growth: extra ions from the solution arrange themselves on the faces of these nuclei, enlarging each crystal.

The relative time available for the second step (growth) depends upon how quickly the water molecules leave the solution.

Case 1 – Fast evaporation (high temperature, blowing air, sun-drying, etc.)

• Water molecules escape quickly; therefore the solution becomes supersaturated in a very short time.
• Large number of nuclei form almost simultaneously because supersaturation is high.
• Each nucleus gets only a small share of the dissolved salt before all liquid disappears.

Hence the individual crystals remain small.

Case 2 – Slow evaporation (room temperature, covered beaker, shaded place, etc.)

• Supersaturation builds up gradually, so fewer nuclei are produced.
• Because water leaves slowly, ions have sufficient time to migrate and add themselves regularly to the existing faces of those few nuclei.

Consequently, each crystal grows larger and well-defined.

Reason summarised mathematically

If $$N$$ is the number of nuclei produced and $$M$$ is the total moles of excess $$\mathrm{NaCl}$$ that must separate, the average moles available per crystal are

$$n_{\text{avg}} = \frac{M}{N}$$

Fast evaporation ⇒ large $$N$$ ⇒ smaller $$n_{\text{avg}}$$ ⇒ small crystals.
Slow evaporation ⇒ small $$N$$ ⇒ larger $$n_{\text{avg}}$$ ⇒ big crystals.

Conclusion

The size of common-salt crystals does change with the rate of evaporation:

• Increasing the rate of evaporation → numerous tiny crystals.
• Decreasing the rate of evaporation → fewer but bigger, well-formed crystals.

Step 1 – Recall the meaning of “immiscible”
Two liquids are said to be immiscible when their molecules do not mix to give a single uniform phase. Each liquid stays as a separate phase because the force of attraction between like molecules (liquid A–A or liquid B–B) is stronger than the attraction between unlike molecules (A–B).

Step 2 – Forces acting inside a separating funnel
When a mixture of two immiscible liquids is poured into a separating funnel and allowed to stand, two phenomena act simultaneously:

• Inter-molecular preference – molecules of the same kind stay together, so the two liquids refuse to mix.
• Gravity – because every point in the liquid experiences its weight, the liquid with the greater density sinks.

Step 3 – Density decides which layer is on top
Assign the densities $$\rho_1$$ and $$\rho_2$$ to the two liquids. If $$\rho_1 > \rho_2$$, liquid 1 is heavier for the same volume, so gravity pulls it to the bottom of the funnel, while the lighter liquid 2 floats above it. The two layers remain distinct because there is practically no mutual solubility.

Result
Thus, in a separating funnel two immiscible liquids arrange themselves in two separate layers: the denser liquid forms the lower layer and the less dense liquid forms the upper layer. Their immiscibility (lack of mutual attraction) prevents them from mixing, and their different densities let gravity stack them one over the other.

Step 1 – Recall the definitions

• Sublimation is the direct change of a substance from the solid state to the vapour (gaseous) state on heating, without passing through the liquid state.
  Example: Dry ice: $$\mathrm{CO_2(s)} \;\longrightarrow\; \mathrm{CO_2(g)}$$
• Evaporation is the slow conversion of the surface molecules of a liquid into vapour at any temperature below its boiling point.
  Example: Water left in an open dish gradually turns into $$\mathrm{H_2O(g)}$$.

Step 2 – State the phase changes involved

Step 3 – Energy considerations

• Both processes require heat energy (endothermic).
• In sublimation the latent heat of fusion is skipped; particles gain enough energy to break all intermolecular bonds at once.
• In evaporation the particles first exist as a liquid and only the more energetic surface molecules escape into the gas phase.

Step 4 – Conditions under which they occur

• Sublimation occurs in solids whose vapour pressure becomes appreciable before melting (e.g. napthalene, iodine, camphor).
• Evaporation can occur for any liquid provided the surrounding air can take up the vapour and the temperature is below the boiling point.

Step 5 – Conclusion

Because the initial states, mechanisms and typical conditions differ, sublimation and evaporation are not the same phenomenon.

Step 1 : Recall the size-ranges and properties of the three kinds of mixtures

Step 2 : Examine the facts about clouds

• A cloud consists of innumerable tiny droplets of liquid water or minute ice crystals dispersed throughout the surrounding air.
• The droplets have an average diameter of roughly $$10\,\mu\text{m}=10\times10^{3}\,\text{nm}$$. (Even though this is slightly above the textbook upper limit of $$1000\,\text{nm}$$, in everyday chemistry such airborne dispersions are still treated as colloids called aerosols.)
• The droplets stay suspended for long periods – they do not settle quickly like the sand particles of a suspension, because they are light and are held up by slow upward air currents.
• Clouds scatter sunlight; the familiar glowing white appearance is nothing but the Tyndall effect on a grand scale.

Step 3 : Match the observations with the definitions

• Not a solution: particles are far larger than $$1\,\text{nm}$$ and do scatter light.
• Not an ordinary suspension: although droplets are large, they remain uniformly spread and do not settle rapidly.
• Most consistent with a colloid: a stable dispersion of a liquid (water) in a gas (air) that exhibits the Tyndall effect. In colloid science such a system is termed a liquid-in-gas aerosol.

Conclusion

Therefore, on the basis of stability and light-scattering behaviour, clouds are best classified as colloidal dispersions (aerosols) of water droplets or ice crystals in air.

Concept recalled — Colloidal dispersion of dust and smoke in air
Air that contains an unusually large number of very fine solid particles (soot, dust) or tiny liquid droplets behaves like a colloid: the gas (air) is the dispersion medium and the solid or liquid particles form the dispersed phase.

The optical property involved — Tyndall effect
When a beam of light passes through a colloidal dispersion, the minute dispersed particles are large enough (roughly 1 nm – 1000 nm) to scatter the incident light in all directions. This phenomenon is called the Tyndall effect.

Step-wise reasoning

1. In a crowded city, exhaust gases from vehicles, factory chimneys and construction activities continuously add smoke and dust to the atmosphere.
2. The suspended particles remain mixed with air for long periods, producing a stable colloidal system.
3. Sunlight (or any bright light) that enters this colloidal atmosphere encounters these particles, and each particle scatters a part of the light.
4. The scattered light reaches our eyes from many random directions instead of only along the straight-line path from the object we are viewing.
5. This additional, randomly scattered light reduces the contrast of distant objects with the background, so they look blurred or faint; the whole scene appears hazy or smoke-filled.

Conclusion
Cities with plenty of smoke and dust look hazy because the suspended particles in air scatter the sunlight (Tyndall effect), making the entire path of light visible and lowering the clarity of distant objects.

Key idea  |   How do we decide?

• Homogeneous mixture (Hm) – only one visible phase; composition and properties are the same throughout. Typical examples are true solutions and solid alloys.
• Heterogeneous mixture (Ht) – shows two or more phases; composition is not uniform. Suspensions, colloids, smokes, fogs, emulsions, etc. fall in this class.

Test every component named in the options.

Check every option against the table.

1. (i) Smoke is marked Hm (should be Ht) → incorrect.
2. (ii) Brass is marked Ht (should be Hm) and muddy water is marked Hm (should be Ht) → incorrect.
3. (iii) Milk is marked Hm (should be Ht) → incorrect.
4. (iv) Every assignment – muddy water (Ht), milk (Ht), blood (Ht), brass (Hm) – matches the table → correct.

Hence, the list in option (iv) is the only one in which all mixtures are classified correctly.

Key concept  — Tyndall effect
When a strong beam of light is passed through a mixture, the path of the light becomes visible only if the dispersed particles are large enough (about 1 nm – 1000 nm) to scatter the light. Such scattering is called the Tyndall effect and it is shown only by colloidal dispersions. It is not observed in:

• true solutions (particle size < 1 nm)
• coarse suspensions whose particles settle quickly and block light instead of scattering it uniformly

We now test each given mixture.

1. Air and dust particles
   Dust particles (diameter ≈ 100 nm to a few µm) are dispersed in air, forming an aerosol, i.e. a colloid. Hence the beam of light becomes visible in a dark room when sunlight enters through a window: the Tyndall effect is observed.
2. Copper sulphate and water
   Copper sulphate dissolves completely, forming a true solution. The ions $$\mathrm{Cu^{2+}}$$ and $$\mathrm{SO_4^{2-}}$$ are about 0.1 nm in size, far below the colloidal range, so no scattering occurs ⇒ no Tyndall effect.
3. Starch and water
   Starch granules do not dissolve molecularly; they swell and remain dispersed with size roughly 10 nm–1000 nm, producing a starch sol (a colloidal solution). Consequently the mixture strongly scatters light ⇒ Tyndall effect is seen.
4. Acetone and water
   Acetone is completely miscible with water and forms a homogeneous true solution (individual molecules, size < 1 nm). Therefore there is no visible scattering ⇒ no Tyndall effect.

Summary of observations

Only mixtures (a) and (c) scatter light, so the correct set of options is (iii) a and c.

Step 1 – Recall the size–based classification of mixtures
A dispersed-phase particle is called

• a solute in a solution if its diameter is smaller than 1 nm
• a suspended particle in a suspension if its diameter is larger than 1000 nm
• a colloidal particle in a colloid if its diameter lies between 1 nm and 1000 nm

Step 2 – Sort every clue from the word-box

• Small-sized particles (< 1 nm), particles remain evenly distributed, transparent, does not settle down, cannot be separated by filtration  ⇒ Solution
• Large-sized particles (> 1000 nm), settles down, separates by filtration, heterogeneous mixture, scatters light  ⇒ Suspension
• Moderate-sized particles (1–1000 nm), does not settle down, particles remain evenly distributed, scatters light, cannot be separated by filtration  ⇒ Colloid

Step 3 – Pick suitable examples from the same word-box

• Solution  ⇒ Salt solution, Brass
• Suspension  ⇒ Sand in water, Mud
• Colloid  ⇒ Milk, Smoke, Butter

Step 4 – Fill Table 5.2 completely

Given data
Volume of the oil pack = 1 L  (= 1000 cm³)
Mass of the oil pack = 910 g

Step 1 : Calculate the density of the cooking oil

The formula for density is $$\rho = \frac{m}{V}$$

Substituting the given values,

$$\rho_{\text{oil}} = \frac{910\;\text{g}}{1000\;\text{cm}^3}$$

$$\rho_{\text{oil}} = 0.91\;\text{g cm}^{-3}$$

Step 2 : Compare the density with that of water

The density of water at room temperature is $$1.0\;\text{g cm}^{-3}$$.

• $$\rho_{\text{oil}} = 0.91\;\text{g cm}^{-3} < \rho_{\text{water}} = 1.0\;\text{g cm}^{-3}$$
• Because the oil is less dense than water and the two liquids are immiscible, they will form two separate layers.

Step 3 : Decide which layer will be on top

Since the oil has lower density, it will float. Therefore,

• Top layer  :  Cooking oil
• Bottom layer  :  Water

Step 4 : Separation method

Use a separating funnel (liquid–liquid separation):

1. Transfer the oil–water mixture into the separating funnel and fit the stopper.
2. Allow the mixture to stand until two clear layers appear.
3. Place a beaker under the funnel. Open the stop-cock to drain the lower (water) layer completely.
4. Close the stop-cock when the oil layer is about to reach the outlet. Collect the oil separately in another beaker.

Step 5 : Diagram to be drawn

• Draw a vertical glass separating funnel with a stop-cock near its tapered end.
• Show two horizontal layers inside: the upper shaded layer labelled “Cooking oil (0.91 g cm⁻³)” and the lower clear layer labelled “Water (1 g cm⁻³)”.
• Draw a stand holding the funnel, a beaker under the stem to collect water, and another beaker kept ready for oil.
• Indicate the direction of flow (arrow) from the stop-cock to the lower beaker.

Thus the immiscible liquids get completely separated.

Step 1 — Recall the Tyndall effect

The Tyndall effect is the visible scattering of a light beam by particles whose sizes are comparable to or larger than the wavelength of visible light (roughly $$400\,\text{nm}$$ to $$700\,\text{nm}$$). Particles must therefore be of at least a few tens of nanometres in diameter to scatter light effectively.

Step 2 — Typical particle sizes in different types of mixtures

Step 3 — Test the Assertion (A)

Because the particles in a true solution are extremely small ($$\lt 1\,\text{nm}$$), they do not scatter light. Hence a beam of light, on passing through a true solution, is not visible. Therefore, Assertion (A) is true.

Step 4 — Test the Reason (R)

Reason (R) claims that “The particles in solutions are larger than $$100\,\text{nm}$$, so they cannot scatter light.” This statement is doubly wrong:

• In reality, the particles in a true solution are smaller than $$1\,\text{nm}$$, not larger than $$100\,\text{nm}$$.
• If particles were indeed larger than $$100\,\text{nm}$$, they would actually scatter light very strongly (as in a colloid or coarse suspension), not fail to do so.

Hence, Reason (R) is false.

Step 5 — Match with the options

Since Assertion (A) is true but Reason (R) is false, the correct choice is

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

Background Concepts

• A suitable separation technique always exploits a difference in some physical property of the components – particle size, solubility, density, volatility, affinity for a stationary phase, etc.
• Whenever we name a method, we must also state which property it makes use of and how that property differs for the two constituents.

Worked explanation for each mixture in Table 5.3

Note on mixtures that cannot be separated by physical means
All seven listed mixtures are physical mixtures, so at least one physical technique exists in each case; hence none is inseparable.

Step 1 · Analyse the information given

• Liquid A boils at $$60\,{}^{\circ}\mathrm{C}$$.
• Liquid B boils at $$90\,{}^{\circ}\mathrm{C}$$.
• Both liquids are miscible.
• The difference in their boiling points is $$90\,{}^{\circ}\mathrm{C}-60\,{}^{\circ}\mathrm{C}=30\,{}^{\circ}\mathrm{C}$$.

NCERT states that if the difference in boiling points of two miscible liquids is more than about $$25\,\text{K}$$, they can be separated by simple distillation (fractional distillation is needed only when the difference is smaller).

Step 2 · Principle of simple distillation

• The liquid with the lower boiling point converts to vapour first.
• The vapour is led through a condenser, where it cools and changes back to liquid.
• The higher-boiling liquid remains in the distillation flask until most of the lower-boiling liquid has been distilled off.

Step 3 · Procedure

1. Pour the mixture of liquids A and B into a round-bottom (distillation) flask and add a few porcelain chips to avoid bumping.
2. Fit the flask with a cork carrying a thermometer so that the bulb is just below the side arm.
3. Connect the side arm to a Liebig condenser inclined slightly downwards toward a clean, dry receiving flask.
4. Gently heat the distillation flask. When the temperature reaches close to $$60\,{}^{\circ}\mathrm{C}$$, liquid A starts boiling and its vapour passes into the condenser.
5. The vapour of A condenses and is collected in the receiver; the temperature remains almost steady near $$60\,{}^{\circ}\mathrm{C}$$ during this period.
6. When no more liquid distils at this temperature, stop heating. Liquid B (with the higher boiling point) is left behind in the distillation flask.

Step 4 · Result

• Liquid A (bp $$60\,{}^{\circ}\mathrm{C}$$) is obtained as the distillate.
• Liquid B (bp $$90\,{}^{\circ}\mathrm{C}$$) remains in the original flask.

Step 5 · Labelled diagram to draw

Draw and label these parts clearly:

• A round-bottom distillation flask containing the mixture and porcelain chips.
• A thermometer inserted through a cork; its bulb just below the side arm.
• A side arm leading to a water-cooled Liebig condenser. Show arrows for water in (nearest the receiver end) and water out (nearest the distillation flask end).
• A receiving flask (conical flask) at the condenser outlet to collect liquid A.
• A stand and clamp holding the apparatus, and a gentle Bunsen burner flame heating the distillation flask.
• Label the temperature range $$\approx60\,{}^{\circ}\mathrm{C}$$ on the thermometer, and indicate vapour of A going into the condenser and residue: liquid B left behind.

With these annotations, the diagram will show how simple distillation separates two miscible liquids whose boiling-point difference is about $$30\,\text{K}$$.

Key idea — all three techniques separate the components of a solution by using the different volatilities of the constituents, but each is carried out in a different way and therefore suits a different purpose.

1. Evaporation

• Apparatus: china dish, Bunsen burner or hot plate, tripod and wire-gauze; no lid or condenser.
• Principle: the solvent (often water) is allowed to evaporate into the surroundings. Only the residue that does not evaporate is obtained.
• Working temperature: close to the solvent’s boiling point; water is steadily lost as vapour.
• Product obtained: generally an amorphous or powdery solid that may still contain soluble impurities because everything that is non-volatile remains behind.
• Uses: obtaining common salt from sea water; concentrating sugar solution before crystallisation.
• Limitations: (i) not suitable if the solid decomposes on strong heating, (ii) solvent is wasted, (iii) purity is limited.

2. Crystallisation

• Apparatus: beaker, glass rod, water-bath, filter paper, watch glass, crystallising dish.
• Principle: prepare a hot, saturated solution. On slow cooling the solute’s solubility falls, so it separates as well-shaped crystals while most impurities stay dissolved in the mother liquor.
• Working temperature: first just below the solvent’s boiling point to make the solution saturated; then allowed to cool undisturbed.
• Product obtained: pure crystalline solid (e.g. blue crystals of $$\mathrm{CuSO_4\,·\,5H_2O}$$).
• Uses: purifying a solid that is heat-stable and more soluble in hot solvent than in cold; preparing sample crystals for X-ray studies, silica gel, alum, etc.
• Limitations: time-consuming; cannot recover the solvent; needs that solubility change with temperature.

3. Distillation

• Apparatus: round-bottom flask, thermometer, condenser, receiver; the setup is closed except for an exit to the receiver.
• Principle: the solution is boiled; the vapour of the more volatile component is led through a condenser, cooled and condensed back to liquid, which is collected separately. Thus both solvent and residue can be obtained.
• Working temperature: exactly at the solvent’s (or lower-boiling liquid’s) boiling point; temperature is monitored by the thermometer.
• Product obtained: pure liquid (distillate) and the less volatile component left behind.
• Uses: (i) obtaining distilled water from tap water, (ii) separating miscible liquids whose boiling points differ by at least $$25\,{}^\circ\mathrm{C}$$ (simple distillation) or less (fractional distillation), (iii) recovering organic solvents such as ethanol.
• Limitations: needs more glassware and heat source; slower than plain evaporation.

4. Side-by-side comparison

5. Which one to prefer?

• Choose evaporation when the only goal is to obtain a solid quickly and the solid does not char; the solvent is cheap or not required (e.g. getting common salt from brine on a large scale).
• Choose crystallisation when high-purity solid crystals are needed and the solute’s solubility changes appreciably with temperature, e.g. purifying impure $$\mathrm{CuSO_4}$$ or sugar.
• Choose distillation when it is necessary to (a) collect the liquid component, (b) keep the solid from decomposing, or (c) separate two miscible liquids with different boiling points, e.g. obtaining distilled water, separating acetone $$\bigl(b.p.=56\,{}^\circ\mathrm{C}\bigr)$$ from ethanol $$\bigl(b.p.=78\,{}^\circ\mathrm{C}\bigr)$$.

Given mixture: sand (insoluble solid), common salt (soluble solid, $$\mathrm{NaCl}$$), and naphthalene (volatile solid that sublimes).

We want each component in the pure, dry form. At every stage only one component must be removed while the others remain unaffected. This fixes an unambiguous order of techniques, explained step-by-step below.

1. Sublimation — separates naphthalene.
   • Put the dry mixture in a china dish.
   • Invert a funnel over it, plug the stem with cotton, and heat gently.
   • Naphthalene sublimes: $$\text{solid} \; \xrightarrow{\text{heat}} \; \text{vapour}$$ and then re-solidifies as a white crystalline layer on the cooler funnel walls.
   • Scrape off and collect the pure naphthalene.

2. Dissolve & filtrate — removes sand from the salt solution.
   • The residue (sand + salt) is transferred to a beaker and stirred with a small quantity of water.
   • Common salt dissolves to give $$\mathrm{NaCl(aq)}$$ but sand remains insoluble.
   • Filter the suspension. Filtration gives:
      Residue on filter paper → dry to obtain pure sand.
      Filtrate (salt solution) → taken to the next step.

3. Evaporation/Crystallisation — obtains common salt.
   • Heat the filtrate gently in an evaporating dish until the volume is reduced and salt crystals start appearing at the edge.
   • Stop heating, cool, and allow crystallisation. Filter, wash and dry the crystals to get pure common salt.

Why no other order works:
If we had added water first, naphthalene (insoluble) would still have to be separated, but sublimation would become cumbersome from a wet mixture. Therefore sublimation must be the very first step, followed by filtration (to remove sand) and finally evaporation/crystallisation (to obtain the dissolved salt).

Hence the correct, complete sequence of separation techniques is:

Sublimation → Filtration → Evaporation / Crystallisation.

Step 1 – Nature of the mixture

• The solution contains two liquids — water and acetone — that are completely miscible (they mix in every proportion without forming layers).

Step 2 – Compare their boiling points

• Normal boiling point of acetone: $$56^{\circ}\mathrm{C}$$
• Normal boiling point of water   : $$100^{\circ}\mathrm{C}$$

The difference is $$100^{\circ}\mathrm{C}-56^{\circ}\mathrm{C}=44^{\circ}\mathrm{C}$$. A gap of more than about $$25^{\circ}\mathrm{C}$$ is considered sufficient for simple distillation.

Step 3 – Principle of distillation

At any given pressure a liquid boils when its vapour pressure equals the external pressure. The liquid with the lower boiling point (higher volatility) reaches this condition first, so it vapourises preferentially.

Step 4 – How separation occurs

1. On gentle heating the thermometer quickly rises to about $$56^{\circ}\mathrm{C}$$. Only acetone boils, producing vapour rich in acetone.
2. The vapour passes into the condenser, cools, and liquefies; pure (or almost pure) acetone drips into the receiver.
3. After most acetone has distilled, the temperature of the residue climbs towards $$100^{\circ}\mathrm{C}$$. Now water begins to boil and can be collected in a separate vessel if required.

Step 5 – Conclusion

The large boiling-point difference allows each component to vaporise at a distinctly different temperature, so distillation cleanly separates the more volatile acetone from water. Therefore distillation is an effective method for this mixture.