NCERT SOLUTIONS • CHAPTER 5

Correct Option: (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

Reasoning:

• Muddy water: Heterogeneous suspension (suspended soil particles).
• Milk: Heterogeneous colloid (emulsion of fat globules in water).
• Blood: Heterogeneous colloid (plasma with blood cells).
• Brass: Homogeneous solid-solid solution (alloy of copper and zinc).

Correct Option: (iii) a and c

Detailed Explanation:

• (a) Air and dust particles (Colloid/Suspension): Dust particles in air scatter light beams, showing the Tyndall effect.
• (c) Starch and water (Colloid): Colloidal starch particles ($1-1000\text{ nm}$) scatter light, showing a visible light path.
• Incorrect Options (b & d): Copper sulfate solution and acetone-water mixture are true homogeneous solutions with particle sizes $<1\text{ nm}$, which are too small to scatter light.

Category	Solution	Suspension	Colloid
Properties	• Homogeneous mixture • Small-sized particles ($<1\text{ nm}$) • Transparent • Does not settle down • Cannot be separated by filtration	• Heterogeneous mixture • Large-sized particles ($>1000\text{ nm}$) • Settles down when left undisturbed • Separates by filtration • Scatters light	• Heterogeneous mixture • Moderate-sized particles ($1-1000\text{ nm}$) • Particles remain evenly distributed • Does not settle down • Scatters light (Tyndall effect) • Cannot be separated by filtration
Examples	Salt solution, Vinegar, Brass	Muddy water, Sand in water	Milk, Smoke, Butter

(i) Cake Recipe Mixture Concentration (% m/m):

$$\text{Total Mass of Mixture} = 75 \text{ g (Sugar)} + 420 \text{ g (Flour)} + 5 \text{ g (Baking Soda)} = 500 \text{ g}$$ $$\text{Sugar Concentration} = \left(\frac{75}{500}\right) \times 100\% = \mathbf{15\%\text{ m/m}}$$ $$\text{Flour Concentration} = \left(\frac{420}{500}\right) \times 100\% = \mathbf{84\%\text{ m/m}}$$ $$\text{Sodium Hydrogencarbonate Concentration} = \left(\frac{5}{500}\right) \times 100\% = \mathbf{1\%\text{ m/m}}$$

(ii) Brass Alloy Masses:

Total mass of brass = $120 \text{ g}$, Copper % = $70\%$, Zinc % = $100 - 70 = 30\%$.

$$\text{Mass of Copper} = 70\% \text{ of } 120 \text{ g} = \frac{70}{100} \times 120 = \mathbf{84\text{ g}}$$ $$\text{Mass of Zinc} = 30\% \text{ of } 120 \text{ g} = \frac{30}{100} \times 120 = \mathbf{36\text{ g}}$$

1. Layer Formation & Top Layer:

Yes, cooking oil and water are immiscible and will form two distinct layers. Density of oil $= \frac{910\text{ g}}{1000\text{ mL}} = 0.91\text{ g/mL}$, which is lower than the density of water ($1.0\text{ g/mL}$). Therefore, cooking oil will float on top of the water layer.

2. Method of Separation:

The mixture is poured into a Separating Funnel. Upon standing, oil forms the upper layer and water forms the lower layer. Opening the stopcock slowly allows the denser lower water layer to drain out completely into a beaker. The stopcock is closed just as the oil reaches the valve, leaving the oil in the funnel.

Correct Option: (iii) A is true, but R is false.

Justification:

• Assertion A is True: True solutions do not scatter light or show the Tyndall effect.
• Reason R is False: Solute particles in true solutions are extremely small (less than 1 nm in diameter, not larger than 100 nm), which is why they cannot scatter visible light rays.

Mixture	Method of Separation	Reason for Selection
Mud from muddy water	Coagulation (with alum) followed by Filtration	Alum clumps fine suspended clay particles so they settle down rapidly and get filtered.
Plasma from blood sample	Centrifugation	Rapid spinning forces denser blood cells to settle at the bottom, leaving liquid plasma at top.
Naphthalene and sand	Sublimation	Naphthalene sublimes directly into gas on heating, leaving non-sublimable sand behind.
Chalk powder and common salt	Dissolution in water, Filtration & Evaporation	Salt is water-soluble while chalk is insoluble. Filtration recovers chalk; evaporating filtrate yields salt.
Common salt and water	Evaporation or Distillation	Water vaporizes on heating, leaving solid salt residue (distillation also recovers pure water).
Oil from water	Separating Funnel	Oil and water are immiscible liquids with different densities (oil floats on water).
Pigments of the flower	Paper Chromatography	Different floral pigments have different solubilities and rates of movement up filter paper.

1. Method Suggested: Simple Distillation

2. Reason:

Liquids A and B are miscible, and the difference in their boiling points ($90^\circ\text{C} - 60^\circ\text{C} = 30^\circ\text{C}$) is greater than $25^\circ\text{C}$.

3. Procedure:

Heat the mixture in a distillation flask fitted with a thermometer. Liquid A ($bp = 60^\circ\text{C}$) vaporizes first, passes through the water condenser, condenses, and collects in the receiver flask as distillate. Liquid B ($bp = 90^\circ\text{C}$) remains behind in the flask.

Technique	Principle	Preferred Situation / Advantage
Evaporation	Solvent vaporizes into air upon heating, leaving non-volatile solid solute behind.	Preferred when only the solid solute needs to be recovered (e.g., obtaining crude salt from seawater).
Crystallization	Solute forms highly pure geometric crystals when a hot saturated solution cools slowly.	Preferred over evaporation when purifying heat-sensitive solids or removing soluble impurities without charring (e.g., pure copper sulfate crystals).
Distillation	Vaporizing a liquid followed by condensing its vapors back into liquid distillate.	Preferred when the liquid solvent must be recovered or when separating two miscible liquids with $\Delta bp \ge 25^\circ\text{C}$ (e.g., acetone-water mixture).

(i) Consequences if blood behaved as a suspension:

Suspension particles settle down under gravity when left undisturbed. If blood were a suspension, blood cells (RBCs, WBCs, platelets) would settle down at the bottom of blood vessels during rest or sleep, causing fatal blood vessel blockages (thrombosis) and heart failure.

(ii) Colloidal Phases of Blood:

• Dispersed Phase: Blood cells (RBCs, WBCs, Platelets) and proteins.
• Dispersion Medium: Liquid Plasma (water with dissolved nutrients & salts).

Correct Sequence of Separation:

1. Step 1 — Sublimation: Heat the solid mixture in a china dish covered with an inverted funnel. Naphthalene sublimes into vapors, condenses on the cool inner funnel walls, and is collected. Sand and salt remain in dish.
2. Step 2 — Dissolution & Filtration: Add water to the remaining sand-salt mixture and stir. Filter the mixture using filter paper. Insoluble sand remains on filter paper as residue.
3. Step 3 — Evaporation / Crystallization: Heat the filtrate (salt solution) to evaporate water, leaving pure common salt crystals behind.

Distillation is effective for separating water and acetone because:

• Both liquids are completely miscible with each other.
• There is a large difference in their boiling points: acetone boils at $56^\circ\text{C}$ while water boils at $100^\circ\text{C}$ ($\Delta bp = 44^\circ\text{C} > 25^\circ\text{C}$).
• When heated to $56^\circ\text{C}$, acetone vaporizes cleanly without boiling water, enabling complete physical separation.

(i) Mass of $\text{KNO}_3$ needed for 50 g water at 40°C:

From Table 5.4, solubility of $\text{KNO}_3$ at $40^\circ\text{C} = 62\text{ g}$ per $100\text{ g}$ water.

$$\text{Mass needed for 50 g water} = \frac{62}{2} = \mathbf{31\text{ g}}$$

(ii) Observation upon cooling $\text{KCl}$ solution:

Solubility of $\text{KCl}$ decreases from $54\text{ g/100g water}$ at $80^\circ\text{C}$ to $\approx 36\text{ g/100g water}$ at $25^\circ\text{C}$. As the solution cools, the excess dissolved $\text{KCl}$ ($54 - 36 = 18\text{ g}$) will precipitate out of solution as solid white KCl crystals.

(iii) Temperature Effect & Salt Comparison:

Solubility of solid salts in water increases with temperature. $\text{KNO}_3$ shows the steepest solubility rise (from 21g to 167g), while $\text{NaCl}$ shows almost negligible change (from 36g to 37g).

(i) Concentration Calculations (% m/m):

• Student A: Mass of solute $= 20\text{ g}$, Mass of solution $= 20 + 80 = 100\text{ g}$
  $$\text{Conc}_A = \left(\frac{20}{100}\right) \times 100\% = \mathbf{20\%\text{ m/m}}$$
• Student B: Mass of solute $= 20\text{ g}$, Mass of solution $= 20 + 100 = 120\text{ g}$
  $$\text{Conc}_B = \left(\frac{20}{120}\right) \times 100\% = \mathbf{16.67\%\text{ m/m}}$$
• Student C: Mass of solute $= 30\text{ g}$, Mass of solution $= 30 + 80 = 110\text{ g}$
  $$\text{Conc}_C = \left(\frac{30}{110}\right) \times 100\% = \mathbf{27.27\%\text{ m/m}}$$

(ii) Most Concentrated Solution:

Student C's solution is the most concentrated ($27.27\%$). It contains the highest ratio of solute mass to total solution mass.

(i) Technique 'S': Simple Distillation

(ii) Apparatus Labels:

• A: Distillation Flask (with thermometer)
• B: Water Condenser (Liebig Condenser)
• C: Receiving Conical Flask (Distillate Receiver)

(iii) Separation Evaluation ($\Delta bp \ge 25^\circ\text{C}$):

• (a) Water ($100^\circ\text{C}$) — Acetone ($56^\circ\text{C}$): $\Delta bp = 44^\circ\text{C} \implies$ Can be separated.
• (b) Water — Salt: Liquid-solid solution $\implies$ Can be separated.
• (c) Acetone ($56^\circ\text{C}$) — Alcohol ($78^\circ\text{C}$): $\Delta bp = 22^\circ\text{C} < 25^\circ\text{C} \implies$ Cannot be separated by simple distillation (requires fractional distillation).
• (d) Sand — Salt: Solid-solid mixture $\implies$ Cannot be separated by distillation.
• (e) Alcohol ($78^\circ\text{C}$) — Chloroform ($61^\circ\text{C}$): $\Delta bp = 17^\circ\text{C} < 25^\circ\text{C} \implies$ Cannot be separated by simple distillation.