Solutions and Their Properties – Long Answer Questions

Medium Level (Application & Explanation)

Q1. Define a solution, solute, and solvent. Give three real-life examples (including chemical formulas where applicable) and classify each as solid in liquid, gas in liquid, or liquid in liquid.

Answer:

A solution is a homogeneous mixture in which the components are uniformly distributed. The substance that is dissolved is the solute, and the substance doing the dissolving is the solvent.
Example 1: Sugar in water (C₁₂H₂₂O₁₁ in H₂O) — solid in liquid. Sugar molecules disperse uniformly in water.
Example 2: Carbon dioxide in soft drink (CO₂ in H₂O) — gas in liquid. Gas is dissolved under pressure and gives fizz when released.
Example 3: Ethanol in water (C₂H₅OH in H₂O) — liquid in liquid. Ethanol and water are completely miscible, forming a single uniform phase.
In each case the mixture is clear, shows no settling, and has uniform composition.

Q2. How can you distinguish a solution, a colloid, and a suspension using simple tests and examples?

Answer:

A solution is clear and transparent, particles are < 1 nm, and it shows no Tyndall effect. Example: salt water. It passes through filter paper and light does not scatter.
A colloid has particle size between 1 nm and 100 nm, appears cloudy or milky, and shows the Tyndall effect (scatters light). Example: milk. Colloidal particles do not settle and cannot be removed by ordinary filtration.
A suspension has large particles (> 100 nm), appears cloudy but settles on standing. Example: sand in water. It can be separated by filtration or decantation.
Simple classroom tests: shine a flashlight (Tyndall effect), pass through filter paper, and let the mixture stand to check settling.

Q3. Explain concentration and calculate the molarity of a solution made by dissolving 10 g of NaCl in water to make 500 mL of solution. (Molar mass of NaCl = 58.5 g mol⁻¹)

Answer:

Concentration measures the amount of solute per unit volume of solution. Molarity (M) is moles of solute per litre of solution.
Step 1: Find moles of NaCl = mass / molar mass = 10 g / 58.5 g mol⁻¹ = 0.171 mol (approx.).
Step 2: Convert volume to litres: 500 mL = 0.500 L.
Step 3: Molarity M = moles / volume(L) = 0.171 mol / 0.500 L = 0.342 M.
So the solution has a molarity of 0.342 M. This tells us the number of NaCl moles dissolved in each litre of solution.

Q4. Describe the principle of dilution and find the volume of 1.0 M hydrochloric acid (HCl) required to prepare 250 mL of 0.10 M HCl.

Answer:

Dilution means reducing concentration by adding more solvent. The number of moles of solute remains the same before and after dilution. The relation is M₁V₁ = M₂V₂.
Given: M₁ = 1.0 M (stock), M₂ = 0.10 M (desired), V₂ = 250 mL = 0.250 L.
Using formula: V₁ = (M₂ × V₂) / M₁ = (0.10 × 0.250) / 1.0 = 0.025 L = 25 mL.
So 25 mL of 1.0 M HCl is taken and diluted with water up to 250 mL to get 0.10 M HCl. Always add acid to water, not water to acid, for safety.

Q5. Explain the terms unsaturated, saturated, and supersaturated solutions with the example of sugar in water and describe how temperature affects solubility.

Answer:

An unsaturated solution contains less solute than the maximum that can dissolve at a given temperature. More sugar will dissolve.
A saturated solution contains the maximum amount of solute that can dissolve at that temperature; any extra solute remains undissolved. For sugar in water at a fixed temperature, crystals appear if you add more.
A supersaturated solution holds more solute than the normal maximum at that temperature; it is unstable and can crystallize on disturbance.
Temperature usually increases the solubility of solids in liquids: warm water dissolves more sugar than cold water. Thus heating can convert saturated to unsaturated, while cooling may cause crystallization and form a supersaturated state.

High Complexity (Analytical & Scenario-Based)

Q6. A student adds lemon juice to warm milk and finds it curdles, forming lumps and a clear liquid. Explain the process using colloid concepts. How can the student separate the curd from the whey?

Answer:

Milk is a colloid with fat and protein particles dispersed in water. Adding acid (lemon) lowers the pH and causes casein proteins to lose charge and aggregate — this is coagulation. The dispersed colloidal particles clump together to form curd, while the remaining liquid is whey.
The process: acid neutralizes charges on protein, particles stick and separate from water. This is used in cheese making.
Separation method: use filtration with a muslin cloth or fine filter to strain curd from whey, or use centrifugation in a lab to speed up separation. The curd remains on the cloth, whey drains off.
This illustrates how changing conditions (pH) can destabilize a colloid.

Q7. You have a mixture containing sand, common salt (NaCl), and water. Describe a step-by-step separation procedure and explain the principles used at each step.

Answer:

Step 1: Filtration — Pour the mixture through filter paper or cloth to separate insoluble sand. Principle: sand particles are large and do not dissolve, so they stay on the filter.
Step 2: Evaporation/Crystallization — Heat the filtrate (salt solution) gently to evaporate water until salt crystals start forming, or leave it to evaporate slowly. Principle: water (solvent) evaporates; the dissolved NaCl becomes supersaturated and crystallizes out.
Step 3: Collection — Collect the salt crystals and dry them. Optionally, condense evaporated water (by distillation) to recover pure water.
These steps use difference in solubility and physical state to separate components.

Q8. Explain why the Tyndall effect occurs in colloids but not in true solutions. Propose a simple classroom experiment using sugar solution and milk to show the difference.

Answer:

The Tyndall effect is light scattering by particles in a mixture. Colloidal particles (1–100 nm) are large enough to scatter light, making a visible beam. In true solutions, particle size is < 1 nm and too small to scatter visible light, so no beam appears.
Classroom experiment: prepare a sugar solution (true solution) and a dilute milk sample (colloid). In a dark room, shine a narrow flashlight or laser pointer through each. The milk will show a visible light path (Tyndall effect), while the sugar solution will not.
Ensure safety with low-power laser. The experiment highlights particle size and optical behavior differences.

Q9. You need to recover pure water from brine (salt water) at home. Describe the process you would use, explain why it works, and mention any energy or efficiency considerations.

Answer:

Use simple distillation: heat brine until water vaporizes, then condense the vapor on a cool surface to collect pure water; salt remains behind. Principle: different boiling points — water boils at 100 °C, leaving non-volatile salt.
Setup: heat brine in a flask, direct steam through a condenser or cooled tube, collect condensate in a clean container. This yields distilled water.
Energy/efficiency: distillation is energy-intensive because heating large volumes takes heat. Heat loss reduces efficiency. Solar stills use sunlight for low-energy distillation but are slower. For home use, small batches are practical; large-scale desalination needs efficient heaters or heat recovery.

Q10. Differentiate between a concentrated and a dilute solution both qualitatively and quantitatively. How do changes in concentration affect some physical properties of a solution?

Answer:

Qualitatively, a concentrated solution has a large amount of solute per unit volume; a dilute solution has a small amount. For example, concentrated sugar water tastes very sweet.
Quantitatively, concentration is measured as molarity (M), percent by mass, or g per L. A solution of 1.0 M NaCl is more concentrated than 0.1 M NaCl.
Changing concentration affects physical properties: higher solute concentration typically raises boiling point and lowers freezing point (colligative effects), and often increases electrical conductivity if the solute is ionic. Viscosity and density also change with concentration.
Thus concentration controls both sensory properties and measurable physical behaviour of solutions.