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Samacheer Kalvi 10th Science Chapter 9 Long Answer Questions — Solutions with Answers

Long answer questions (5 marks) from Samacheer Kalvi Class 10 Science Chapter 9 Solutions with detailed answers covering types of solutions, concentration, osmosis, and colligative properties.

Updated 25 May 20268 min read

Chapter 9 — Solutions: Long Answer Questions


1. What is a solution? Classify solutions with examples.
Answer

A solution is a homogeneous mixture of two or more substances. The substance present in larger amount is the solvent; the substance present in smaller amount is the solute.

Classification of solutions by physical state:

TypeSoluteSolventExample
Gaseous solutionGasGasAir (O₂ in N₂), natural gas mixtures
LiquidGasWater vapour in air (humidity)
SolidGasCamphor vapour in air
Liquid solutionGasLiquidAerated drinks (CO₂ in water), O₂ in water
LiquidLiquidEthanol in water, vinegar (acetic acid in water)
SolidLiquidSalt in water, sugar in water
Solid solutionGasSolidHydrogen in palladium (used in H₂ storage)
LiquidSolidMercury in gold (amalgam)
SolidSolidAlloys: brass (zinc in copper), steel (carbon in iron)

Based on amount of solute:

  • Unsaturated: Can dissolve more solute at given temperature
  • Saturated: Maximum amount of solute dissolved; addition of more solute causes precipitation
  • Supersaturated: Contains more solute than a saturated solution (unstable — achieved by cooling a hot saturated solution carefully)

2. What factors affect the solubility of a substance? Explain each factor.
Answer

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature and pressure to form a saturated solution.

Factors Affecting Solubility

1. Nature of solute and solvent (Like dissolves like):
Answer
  • Polar solvents (water) dissolve polar solutes and ionic compounds (NaCl, KNO₃, HCl)
  • Non-polar solvents (benzene, hexane) dissolve non-polar solutes (fats, oils, waxes)
  • Example: NaCl is highly soluble in water but insoluble in benzene
  • Example: Iodine is slightly soluble in water but very soluble in hexane
2. Temperature:
Answer
  • For most solid solutes in liquid solvents: Solubility increases with temperature (e.g., KNO₃, NaCl, sugar) — dissolving is usually endothermic; heating favours dissolution
  • Exception: Some salts show decreased solubility with temperature (e.g., Na₂SO₄, CaCO₃) — dissolving is exothermic
  • For gases in liquids: Solubility decreases with temperature — this is why warm carbonated drinks fizz less
3. Pressure (mainly affects gases — Henry's Law):
Answer
  • Henry's Law: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid (at constant temperature)
  • C = kH × P
  • Examples: CO₂ under pressure in a sealed can; O₂ dissolved in water at higher pressure in scuba diving tanks
  • Application: Soda bottles — CO₂ remains dissolved under pressure; fizzes when opened (pressure decreases)
  • Caution: Deep-sea divers must decompress slowly to avoid nitrogen bubbles forming in blood (bends/decompression sickness)
4. Effect of common ion:
Answer
  • Adding a salt that shares an ion with a slightly soluble salt reduces its solubility (common-ion effect)
  • Example: Adding NaCl reduces solubility of AgCl (both have Cl⁻)

3. Define molarity, mass percentage, and molality. Solve a problem for each.

Mass Percentage (w/w %)

Definition: Mass of solute per 100 g of solution.

Mass % = (mass of solute / mass of solution) × 100 = (mass of solute / (mass of solute + mass of solvent)) × 100

Example: 25 g of NaCl dissolved in 225 g of water.

  • Mass of solution = 25 + 225 = 250 g
  • Mass % NaCl = (25/250) × 100 = 10%

Molarity (M)

Definition: Number of moles of solute per litre of solution.

M = n / V(L) = (mass of solute / molar mass) / V(L)

Unit: mol/L or mol dm⁻³

Example: 5.85 g of NaCl (molar mass = 58.5 g/mol) dissolved to make 500 mL of solution.

  • n(NaCl) = 5.85/58.5 = 0.1 mol
  • V = 500 mL = 0.5 L
  • M = 0.1/0.5 = 0.2 M

Molality (m)

Definition: Number of moles of solute per kilogram of solvent.

m = n / mass of solvent (kg)

Unit: mol/kg

Advantage over molarity: Molality is independent of temperature (uses mass, not volume). Molarity changes slightly with temperature.

Example: 10 g of glucose (C₆H₁₂O₆, molar mass = 180 g/mol) dissolved in 500 g of water.

  • n(glucose) = 10/180 = 0.0556 mol
  • Mass of solvent = 500 g = 0.5 kg
  • m = 0.0556/0.5 = 0.111 mol/kg

4. Explain osmosis and osmotic pressure. Give examples from daily life and biology.

Osmosis

Osmosis is the spontaneous net movement of solvent molecules through a semi-permeable membrane from a region of lower solute concentration (higher solvent concentration) to a region of higher solute concentration (lower solvent concentration).

Semi-permeable membrane: A membrane that allows solvent molecules to pass through but not solute particles. Examples: cell membranes, egg membrane, parchment paper, RBC membrane, dialysis tubing.

Driving force: The difference in solvent chemical potential (or water potential) across the membrane.


Osmotic Pressure (π)

The minimum pressure that must be applied to the solution to just stop the osmosis of solvent from pure solvent into the solution.

Van't Hoff equation:

π = MRT = (n/V)RT

Where M = molarity, R = gas constant, T = temperature in K

Higher solute concentration → higher osmotic pressure.

Types of solutions with respect to a biological cell:

  • Isotonic: Same concentration as cell contents — cell maintains normal size
  • Hypotonic: Less concentrated than cell — water enters cell by osmosis → cell swells → may burst (lysis in animal cells; turgidity in plant cells)
  • Hypertonic: More concentrated than cell — water leaves cell by osmosis → cell shrinks (plasmolysis in plants; crenation in animal cells)

Daily Life and Biological Examples of Osmosis

ExampleExplanation
Salting vegetables and meatSalt draws out water by osmosis (food preservation — bacteria dehydrate)
Plants absorbing water from soilSoil water (hypotonic) moves into root cells (hypertonic) by osmosis
Wilting of plantsLoss of water (drought/high soil salinity) → cells lose turgidity → wilting
Rehydration salts (ORS)Balanced ionic concentration helps intestine absorb water by osmosis
Kidney dialysisArtificial semi-permeable membrane removes waste from blood
Reverse osmosis (RO) water purifiersHigh pressure applied to force water through membrane against osmotic gradient — removes dissolved salts
Preserving food in jam/jellyHigh sugar concentration draws water out of microbes by osmosis — preserves food

5. What are colligative properties? Explain boiling point elevation and freezing point depression.
Answer

Colligative properties are properties of solutions that depend only on the number of solute particles (moles) dissolved, not on the nature or identity of the solute.

The four colligative properties are: 1. Lowering of vapour pressure 2. Elevation of boiling point 3. Depression of freezing point 4. Osmotic pressure


Boiling Point Elevation (ΔTb)

When a non-volatile solute is dissolved in a solvent, the boiling point of the solution is higher than that of the pure solvent.

Reason: The vapour pressure of the solution is lower than pure solvent (Raoult's Law). More heat must be supplied to raise vapour pressure to atmospheric pressure → higher boiling point.

Formula
ΔTb = Kb × m

Where Kb = ebullioscopic constant of solvent, m = molality

Practical example: Adding salt to cooking water raises the boiling point → food cooks faster (though the increase is very small — about 0.5°C per mole per kg for water).


Freezing Point Depression (ΔTf)

When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.

Reason: Solute particles interfere with the formation of the ordered crystal lattice of the solid → lower temperature needed to freeze.

Formula
ΔTf = Kf × m

Where Kf = cryoscopic constant of solvent, m = molality

Practical examples:

  • Antifreeze in car radiators: Ethylene glycol dissolved in water lowers the freezing point below −40°C → prevents radiator from freezing in winter
  • Salting roads in winter: Salt (NaCl) dissolves on wet roads, depressing the freezing point → prevents ice formation
  • Sea water: Dissolved salts lower the freezing point to about −1.8°C → ocean freezes at a lower temperature than fresh water

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Frequently Asked Questions

Molarity (M) is the number of moles of solute dissolved in 1 litre (1000 mL) of solution. Formula: M = n/V = (mass of solute / molar mass of solute) / volume of solution in litres. Unit: mol/L or M. Example: Dissolving 40 g of NaOH (molar mass = 40 g/mol) in enough water to make 500 mL solution: M = (40/40) / 0.5 = 2 M.