Solutions-NCERT-Class-9-Science (Exploration)-Chapter-5-Exploring Mixtures and their Separation-CBSE

• Muddy water is a suspension, which is a heterogeneous mixture containing relatively large solid particles (greater than 1000 nm in diameter). Because these particles are large and heavy, gravity overcomes their thermal kinetic motion, causing them to settle at the bottom over time.
• In contrast, milk is a colloid (an emulsion of fat droplets in water). The dispersed fat particles in milk are much smaller (1 nm to 1000 nm) and remain suspended continuously due to rapid Brownian motion and electrostatic repulsion, preventing them from settling out.

Evaporation and boiling are both vaporisation processes, but they differ fundamentally in the following ways:

• Surface vs. Bulk Phenomenon: Evaporation is a surface phenomenon occurring only at the liquid surface, whereas boiling is a bulk phenomenon occurring throughout the entire liquid volume.
• Temperature Requirement: Evaporation occurs naturally at any temperature below the boiling point, whereas boiling takes place only at a specific fixed temperature (the boiling point).
• Rate & Bubbles: Evaporation is a slow process without bubble formation. Boiling is a rapid process characterized by the formation and movement of vapour bubbles throughout the liquid.

This phenomenon occurs due to the Tyndall Effect.

• The atmosphere in a dense forest contains mist, fog, and suspended dust particles, which act as colloidal particles dispersed in air.
• When sunlight passes through small gaps between leaves, these tiny particles scatter the light rays in all directions, making the path of the light beam clearly visible to our eyes.

The mixture of oil and water is a heterogeneous mixture.

Reason: Oil and water are immiscible liquids. They do not dissolve in each other and form two visibly distinct layers with a sharp boundary, maintaining a non-uniform composition throughout.

Other Examples of Heterogeneous Mixtures:

• Sand stirred in water
• Chalk powder suspended in water
• Sawdust mixed with water
• Smoke (solid carbon particles suspended in air)

Formula: Mass % (m/m) = \(\frac{\text{Mass of solute}}{\text{Mass of solution}}\)  × 100

Step 1: Identify given quantities.

• Mass percentage (% m/m) of Zinc Oxide = 4%
• Total mass of talcum powder (solution/mixture) = 300 g

Step 2: Apply the Mass by Mass Percentage Formula.

• % m/m = (Mass of solute / Total mass of mixture) × 100

Step 3: Substitute values and calculate.

• 4 = (Mass of Zinc Oxide / 300 g) × 100
• Mass of Zinc Oxide = (4 × 300) / 100 = 12 g

Conclusion: There are 12 grams of zinc oxide present in 300 g of the talcum powder.

Formula: Volume % (v/v) = \(\frac{\text{Volume of solute}}{\text{Volume of solution}}\)  × 100

Step 1: Calculate the total volume of solute (orange juice concentrate).

• Volume of 1 tablespoon = 15 mL
• Volume of 2 tablespoons (solute) = 2 × 15 mL = 30 mL

Step 2: Identify the total volume of solution.

• Total volume of orange juice per glass (solution) = 150 mL

Step 3: Apply the Volume by Volume Percentage Formula.

• % v/v = (Volume of solute / Total volume of solution) × 100
• % v/v = (30 mL / 150 mL) × 100 = (1 / 5) × 100 = 20%

Conclusion: The concentration of orange juice concentrate in the prepared drink is 20% v/v.

Step 1: Understand the target concentration.

• Commercial vinegar requires a 5% v/v concentration of acetic acid in water.

Step 2: Calculate required volumes for standard 100 mL batch.

• To make 100 mL of 5% v/v vinegar, you need 5 mL of pure (glacial) acetic acid.

Step 3: Procedure for preparation.

1. Measure accurately 5 mL of liquid glacial acetic acid using a calibrated pipette.
2. Transfer it into a 100 mL volumetric flask containing approximately 50 mL of distilled water.
3. Mix gently and add water until the total volume reaches the 100 mL mark.

General Rule: Mix 5 parts by volume of glacial acetic acid with 95 parts by volume of water.

(i) Answer: less than

• Explanation: As temperature increases from 20 °C to 60 °C, the solubility curve for Compound A rises, showing higher solubility at 60 °C.

(ii) Answer: less than

• Explanation: The solubility curve for Compound B rises significantly as temperature increases from 20 °C to 60 °C.

(iii) Answer: Compound 'B' ; Compound 'A'

• Explanation: Compound B's curve has a much steeper upward slope (rising from ~100 g to 350 g/100 g water) than Compound A's curve, indicating a far greater increase in solubility per degree rise in temperature.

Compound 'B' will deposit significantly more solid.

Reasoning & Analysis:

• The solubility of a substance represents the maximum amount that can remain dissolved in 100 g of solvent at a given temperature.
• Between 80 °C and 60 °C, the solubility of Compound B drops rapidly (steep decrease on graph).
• Because Compound B's solubility decreases by a much larger margin than Compound A's over the same temperature interval, the excess solute that can no longer stay dissolved precipitates out as crystals, depositing a greater mass of solid Compound B.

Yes, the rate of evaporation directly determines the size of the resulting salt crystals.

Slower Evaporation (Decreased Rate): When water evaporates slowly (e.g., at room temperature), solute particles have ample time to migrate and lock into organized crystalline lattice positions.

This slow, undisturbed growth results in larger, well-shaped, shiny crystals.

Faster Evaporation (Increased Rate): Rapid heating or high evaporation forces fast precipitation of solute particles.

The particles assemble rapidly and haphazardly, giving rise to numerous smaller, finer crystals.

(i) True.

• Explanation: Evaporation leaves solid salt in the dish, while distillation recovers pure water distillate while leaving salt behind in the flask.

(ii) False.

• Correction: Distillation requires a significant difference in boiling points (at least 25 °C) between miscible liquids so that one vaporises distinctly before the other.

(iii) False.

• Correction: In paper chromatography, the solvent level MUST be below the sample spot so that the sample is not washed directly into the solvent pool, but instead travels up the paper via capillary action.

(iv) False.

• Correction: Evaporation drives off liquid completely to leave crude/amorphous residue, whereas crystallization forms highly pure, geometrically structured crystals from a cooled saturated solution.

Reason 1: Chemical Immiscibility

• Immiscible liquids (like mustard oil and water) have different molecular polarities and cannot dissolve in each other.

Reason 2: Density Difference

• Driven by gravity, the denser liquid (water, density ~1.0 g/cm³) settles at the bottom, while the less dense liquid (mustard oil, density ~0.9 g/cm³) floats to the top, creating two distinct liquid layers.

Yes, sublimation is distinctly different from evaporation:

• Sublimation is the direct transition of a solid into gas/vapour phase (and vice-versa upon cooling as deposition) without passing through an intermediate liquid state.
• Evaporation is the transition of a liquid into vapour phase occurring at temperatures below its boiling point.

Examples:

• Sublimation: Heating camphor, ammonium chloride, or naphthalene.
• Evaporation: Drying wet clothes or recovering salt from seawater.

Clouds are colloids (specifically aerosols).

Justification:

• Particle Size: The tiny water droplets or ice crystals have diameters within the colloidal range (1 nm to 1000 nm).
• Non-settling Behavior: They do not settle rapidly under gravity (unlike suspensions), but remain uniformly dispersed throughout the atmospheric air medium.
• Light Scattering: Clouds scatter sunlight efficiently (Tyndall Effect), making them appear white or gray.

The hazy appearance is caused by the Tyndall Effect.

Explanation: Smoke and dust particles suspended in urban air act as colloidal particles dispersed in a gas medium. When sunlight passes through this polluted air, these microscopic particles scatter light in all directions, obscuring distant objects and creating a visible, cloudy haze.

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

Why the others are wrong:

(i) smoke is heterogeneous (solid particles in air);

(ii) brass is homogeneous (an alloy), vinegar is homogeneous, muddy water is heterogeneous;

(iii) milk is not a solution.

Correct option : (iii) a and c

Why the others are incorrect ?:

(a) air with dust particles scatters light, and (c) starch in water is a colloid, so it scatters light.

(b) copper sulfate solution and (d) acetone–water are true solutions with particles < 1 nm: no scattering.

So (i), (ii) and (iv) are incorrect.

1. Calculate Total Mass of Solution/Mixture:

Total Mass = Mass(Sugar) + Mass(Flour) + Mass(Sodium Hydrogencarbonate)

Total Mass = 75 g + 420 g + 5 g = 500 g

2. Apply Mass Percentage Formula (% m/m):

% m/m = \(\frac{\text{Mass of component}}{\text{Mass of mixture}}\) × 100

3. Calculate Individual Concentrations:

• Sugar (% m/m) = \(\frac{75\,g}{500\,g}\) × 100 = 15.0% m/m
• All-purpose Flour (% m/m) = \(\frac{420\,g}{500\,g}\) × 100 = 84.0% m/m
• Sodium Hydrogencarbonate (% m/m) = \(\frac{5\,g}{500\,g}\) × 100 = 1.0% m/m

Verification: 15.0% + 84.0% + 1.0% = 100.0%

Identify Given Data:

• Total mass of Brass alloy = 120 g
• % m/m of Copper (Cu) = 70%

Calculate Mass of Copper:

• Mass of Copper = \(\frac{70}{100}\) × 120 g = 84.0 g

Calculate Mass of Zinc (Zn):

• Method A (% remaining): % Zinc = 100% - 70% = 30%

Mass of Zinc = \(\frac{30}{100}\) × 120 g = 36.0 g

• Method B (Difference): Mass of Zinc = Total Mass - Mass of Copper = 120 g - 84 g = 36.0 g

Final Answer: Copper = 84 g, Zinc = 36 g

Layer Formation: Yes, oil and water are immiscible liquids and will form two separate, distinct layers when mixed.

Top Layer Determination:

• Density of Water = 1.00 g/mL (or 1000 g/L).
• Density of Cooking Oil = \(\frac{mass}{volume}\) = \(\frac{910}{1000}\) = 0.91 g/mL.
• Since 0.91 g/mL < 1.00 g/mL, cooking oil is less dense than water and will float on top (forming the upper layer).

Separation Method & Procedure:

Apparatus: Separating Funnel.

Procedure: Pour the oil-water mixture into a separating funnel and let it stand undisturbed until two clear layers form. Open the stopcock at the bottom of the funnel to carefully drain the denser bottom water layer into a receiving beaker. Close the stopcock immediately when the interface reaches the stopcock, leaving pure oil inside the funnel.

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

Detailed Scientific Analysis:

• Assertion (A) is TRUE: True solutions do not scatter light, so the path of a laser beam remains invisible inside the solution.
• Reason (R) is FALSE: Solute particles in a true solution are extremely small — less than 1 nanometre (< 1 nm or 10⁻⁹ m) in diameter — NOT larger than 100 nm. Because their dimensions are far smaller than the wavelength of visible light, they cannot scatter light rays.

Suggested Method: Simple Distillation

Justification: The boiling point difference between Liquid A (60 °C) and Liquid B (90 °C) is ΔT = 90 °C - 60 °C = 30 °C. Since the difference exceeds the threshold of 25 °C, simple distillation effectively separates them without requiring a fractionating column.
Process & Apparatus Setup:

1. The mixture is placed in a round-bottom distillation flask fitted with a thermometer and connected to a Liebig water condenser.
2. Upon heating, Liquid A (lower B.P. 60 °C) vaporizes first at 60 °C. Its vapours rise, pass through the water-cooled condenser, condense back into liquid distillate, and collect in a receiving flask.
3. Liquid B (higher B.P. 90 °C) remains unchanged in the distillation flask until Liquid A is completely separated.

Comparison of Evaporation, Crystallization, and Distillation :

(i) If blood behaved like a true suspension inside the body:

• In a suspension, large particles settle down at the bottom due to gravity whenever the system is left undisturbed.
• If blood were a suspension, blood cells (RBCs, WBCs, platelets) would settle out of plasma inside blood vessels whenever a person sleeps or sits still.
• This settling would clog arteries and veins, stop blood circulation, prevent oxygen delivery to vital organs, and cause fatal blockages.

(ii) Identification of Colloidal Components in Blood:

• Dispersed Phase (Solid components): Red Blood Cells (RBCs), White Blood Cells (WBCs), and Platelets.
• Dispersion Medium (Liquid medium): Blood Plasma (aqueous solution containing dissolved proteins, electrolytes, glucose, and waste).

Correct Sequence of Separation Techniques:

Step 1 – Sublimation: heat the mixture in a china dish under an inverted, cotton-plugged funnel; naphthalene sublimes and deposits on the funnel.

Step 2 – Dissolving (with heating): add water to the residue (sand + salt) and stir/heat; salt dissolves, sand does not.

Step 3 – Filtration: filter; sand stays on the filter paper, salt solution is the filtrate. (Evaporating the filtrate then gives salt.)

Distillation is highly effective for separating water and acetone because:

• Miscibility: Acetone and water dissolve completely in each other to form a single homogeneous liquid phase.
• Large Boiling Point Difference: Acetone boils at 56 °C, whereas Water boils at 100 °C. The boiling point difference ΔT = 100 °C - 56 °C = 44 °C, which is well above the required 25 °C threshold for simple distillation.
• Clean Separation: Upon heating to ~56 °C, acetone vaporizes completely while water stays liquid. Condensed acetone is collected as pure distillate.

Part (i): Saturated Solution of Potassium Nitrate at 40°C in 50 g Water

• From Table, solubility of Potassium Nitrate (KNO₃) at 40 °C = 62 g per 100 g water.
• Mass of KNO₃ needed for 50 g water = (62 g / 100 g) × 50 g = 31.0 g.

Part (ii): Cooling Saturated Potassium Chloride Solution from 80°C to Room Temperature

• At 80 °C, solubility of KCl = 54 g per 100 g water.
• At room temperature (25 °C), solubility of KCl = ~36.2 g per 100 g water.

Observation & Explanation: As the solution cools from 80 °C to 25 °C, water can no longer hold 54 g of dissolved KCl. The excess dissolved salt (~17.8 g per 100 g water) will separate out of the liquid phase and form solid shiny crystals of potassium chloride (Crystallization).

Part (iii): Temperature Effect & Comparative Salt Analysis (10°C to 80°C)

General Effect: The solubility of solid salts in water generally increases with an increase in temperature.

Comparative Increase (10 °C → 80 °C):

1. Potassium Nitrate (KNO₃): Massive increase from 21 g to 167 g (Δ = +146 g / 100 g water) — most temperature-sensitive.
2. Ammonium Chloride (NH₄Cl): Substantial increase from 24 g to 66 g (Δ = +42 g / 100 g water).
3. Potassium Chloride (KCl): Moderate increase from 35 g to 54 g (Δ = +19 g / 100 g water).
4. Sodium Chloride (NaCl): Minimal increase from 36 g to 37 g (Δ = +1 g / 100 g water) — practically independent of temperature.

Part (i): Mass Percentage (% m/m) Calculations

Student A (20 g sugar in 80 g water):
   • Total mass = 20 g + 80 g = 100 g
   • % m/m = \(\frac{20}{100}\) × 100 = 20.0% m/m
Student B (20 g sugar in 100 g water):
   • Total mass = 20 g + 100 g = 120 g
   • % m/m = \(\frac{20}{120}\) × 100 = 16.67% m/m
Student C (30 g sugar in 80 g water):
   • Total mass = 30 g + 80 g = 110 g
   • % m/m = \(\frac{30}{110}\) × 100 = 27.27% m/m

Part (ii): Most Concentrated Solution Determination

Student C's solution is the most concentrated (27.27% m/m).

Explanation: Concentration measures the mass of solute per unit total mass of solution. Student C dissolved the largest amount of solute (30 g) in the smallest relative amount of solvent (80 g), resulting in the highest proportion of dissolved sugar compared to A (20.0%) and B (16.67%).

(i) Technique Marked as 'S': Simple Distillation.
(ii) Labelled Apparatus Components:
   • A: Thermometer
   • B: Water Condenser (Liebig Condenser)
   • C: Conical Flask (Receiving Flask / Distillate Collector)
(iii) Mixture Selection for Simple Distillation (using Table BP data):
Simple distillation requires either a dissolved solid in liquid or two miscible liquids with a boiling point difference ΔT ≥ 25 °C.
   • (a) Water (100 °C) — Acetone (56 °C): ΔT = 44 °C (≥ 25 °C) → YES, CAN BE SEPARATED BY DISTILLATION S.
   • (b) Water — Salt: Salt is a non-volatile solid, water boils at 100 °C → YES, CAN BE SEPARATED BY DISTILLATION S.
   • (c) Acetone (56 °C) — Alcohol (78 °C): ΔT = 22 °C (< 25 °C) → NO (requires fractional distillation).
   • (d) Sand — Salt: Both are solids → NO (requires dissolution, filtration, evaporation).
   • (e) Alcohol (78 °C) — Chloroform (61 °C): ΔT = 17 °C (< 25 °C) → NO (requires fractional distillation).
   • (f) Alcohol (78 °C) — Benzene (80 °C): ΔT = 2 °C (< 25 °C) → NO (requires fractional distillation).
Final Answer for (iii): Mixtures (a) water–acetone and (b) water–salt.