CBSE Important Questions Class 12 Chemistry Chapter 1 – Solutions

Class 12 Chemistry Chapter 1 Solutions is the first chapter of the CBSE Class 12 Chemistry syllabus. A solution is a homogeneous mixture of two or more substances. Students learn the different ways of expressing concentration, the solubility of solids and gases in liquids, Henry's law, Raoult's law, ideal and non-ideal solutions, and the four colligative properties used to find molar mass.

These CBSE important questions for Class 12 Chemistry Chapter 1 cover Henry's law applied to carbonated drinks, the condition for an ideal solution, types of solutions, the role of the semi-permeable membrane in osmosis, isotonic and hypertonic solutions, intermolecular forces in different pairs, and calculations on molarity, molality and mole fraction. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. A soft drink company is testing two gases, CO₂ and N₂, for carbonating their beverages. The Henry's law constant (Kₕ) for CO₂ is much lower than that of N₂ at the same temperature. The quality engineer plots partial pressure against mole fraction for both gases. Identify which line is CO₂ and which is N₂, and predict which gas dissolves more readily at the same partial pressure.

Graph of partial pressure versus mole fraction for two gases with different slopes

Answer: According to Henry's law, p = Kₕ x, where p is the partial pressure and x is the mole fraction of the gas in solution.
The graph of p against x is a straight line passing through the origin, and its slope is Kₕ.
• CO₂ has a lower Kₕ, so it gives the line with the smaller slope.
• N₂ has a higher Kₕ, so it gives the steeper line.
Which dissolves more readily? Rearranging Henry's law, x = p/Kₕ. For the same partial pressure, a smaller Kₕ gives a larger mole fraction.
So CO₂ dissolves more readily in the beverage. This is exactly why fizzy drinks are carbonated with CO₂ and not with N₂.

Q2. The formation of a solution from two components can be considered in three steps:
(i) Pure solvent → separated solvent molecules, ΔH₁
(ii) Pure solute → separated solute molecules, ΔH₂
(iii) Separated solvent and solute molecules → solution, ΔH₃
The solution so formed will be ideal if
(a) ΔHsoln = ΔH₃ − ΔH₁ − ΔH₂
(b) ΔHsoln = ΔH₁ + ΔH₂ + ΔH₃
(c) ΔHsoln = ΔH₁ + ΔH₂ − ΔH₃
(d) ΔHsoln = ΔH₁ − ΔH₂ − ΔH₃

Answer: (b) ΔHsoln = ΔH₁ + ΔH₂ + ΔH₃
The total enthalpy change of forming the solution is simply the sum of the enthalpy changes of the three steps.
For an ideal solution, ΔHsoln = 0 and ΔVmixing = 0, which means the solute-solvent interactions are exactly the same in strength as the solute-solute and solvent-solvent interactions.

Q3. Write one example each of a solid in gas and a liquid in gas solution.

Answer: Solid in gas: camphor in nitrogen gas (also iodine vapour in air)
Liquid in gas: chloroform mixed with nitrogen gas (also water vapour in air, which is humidity)
A solution is any homogeneous mixture, so it need not be a liquid. There are nine possible types of solution, depending on whether the solute and solvent are solid, liquid or gas.

Q4. Why is a semi-permeable membrane so important in osmosis? What are isotonic, hypotonic and hypertonic solutions? Does osmosis take place in all three?

Answer: A semi-permeable membrane is important because it allows only the solvent molecules to pass through, not the solute. This one-way selectivity is what makes osmosis possible.
• Isotonic solutions: two solutions having the same osmotic pressure at a given temperature.
• Hypotonic: a solution with lower osmotic pressure than the one on the other side.
• Hypertonic: a solution with higher osmotic pressure than the one on the other side.
Does osmosis take place in all three? No. Osmosis takes place only between hypotonic and hypertonic solutions. Between isotonic solutions there is no net flow, because the osmotic pressures are already balanced.

Q5. List three general applications of the phenomenon of osmosis.

Answer: 1. Shrinking and swelling of blood cells when placed in a salt solution more or less concentrated than 0.9% NaCl. In a hypertonic solution the cells shrink, and in a hypotonic solution they swell and may burst.
2. Movement of water from the soil into plant roots, which is how plants draw up water.
3. Preservation of meat and fruit by salting or by adding sugar. The high salt or sugar concentration draws water out of bacterial cells, killing them and preventing spoilage.

Q6. Suggest the most important type of intermolecular attractive interaction in each of the following pairs.
(i) n-hexane and n-octane
(ii) I₂ and CCl₄
(iii) NaClO₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆O)

Answer: (i) Both are non-polar, so the interaction is London dispersion forces.
(ii) Both are non-polar, so again London dispersion forces.
(iii) NaClO₄ gives Na⁺ and ClO₄⁻ ions in solution, and water is polar. The interaction is ion-dipole.
(iv) Both are polar molecules, so dipole-dipole interactions.
(v) Both are polar molecules, so dipole-dipole interactions.

Q7. Define mole fraction, molality, molarity and mass percentage.

Answer: • Mole fraction (x): the number of moles of one component divided by the total number of moles of all components. It has no unit, and the mole fractions of all components add up to 1.
• Molality (m): the number of moles of solute per kilogram of solvent. Its unit is mol/kg. It does not change with temperature, since mass does not change with temperature.
• Molarity (M): the number of moles of solute per litre of solution. Its unit is mol/L. It does change with temperature, because volume changes with temperature.
• Mass percentage: (mass of solute ÷ mass of solution) × 100

Q8. The molarity and molality of a solution at a certain temperature were recorded. Identify the correct values.
(a) Molarity = 0.40 M and molality = 0.49 m
(b) Molarity = 0.59 M and molality = 0.70 m
(c) Molarity = 0.50 M and molality = 0.49 m
(d) Molarity = 0.79 M and molality = 0.79 m

Answer: (c) Molarity = 0.50 M and molality = 0.49 m
For a dilute aqueous solution, molality and molarity are close in value but not equal. Molality is based on the mass of the solvent while molarity is based on the volume of the solution, which also contains the solute.
Option (c) shows the two values close to each other with molality slightly less, which matches a typical dilute aqueous solution.

Q9. For a solution, the mole fraction of the solute is 0.5 and the number of moles of solute is 6.5 mol. The number of moles of solvent is
(a) 4.5
(b) 6.5
(c) 9.5
(d) 3.5

Answer: (b) 6.5
Mole fraction of solute = moles of solute ÷ total moles
0.5 = 6.5 / (6.5 + nsolvent)
6.5 + nsolvent = 13
nsolvent = 6.5 mol
This makes sense: when the mole fraction is exactly 0.5, the two components are present in equal numbers of moles.

Q10. A solution of glucose in water is labelled 10% w/w. Find the molality and the mole fraction of each component. If the density of the solution is 1.2 g/mL, what is the molarity?

Answer: Take 100 g of solution, so glucose = 10 g and water = 90 g. (Molar mass of glucose = 180 g/mol)
Moles of glucose = 10/180 = 0.0555 mol
Moles of water = 90/18 = 5 mol
Molality = 0.0555 / 0.090 kg = 0.617 m
Mole fraction of glucose = 0.0555 / (0.0555 + 5) = 0.011
Mole fraction of water = 1 − 0.011 = 0.989
Molarity: Volume of solution = mass/density = 100/1.2 = 83.33 mL = 0.08333 L
Molarity = 0.0555 / 0.08333 = 0.67 M

Q11. How does an increase in temperature affect the solubility in endothermic and exothermic dissolution?

Answer: This follows from Le Chatelier's principle.
• Endothermic dissolution (ΔHsoln is positive, heat is absorbed): raising the temperature pushes the equilibrium forward, so solubility increases.
• Exothermic dissolution (ΔHsoln is negative, heat is released): raising the temperature pushes the equilibrium backward, so solubility decreases.
This is why most salts dissolve better in hot water, while gases (whose dissolution is exothermic) dissolve less in hot water.

Q12. What is meant by the concentration of a solution? Name the different ways of expressing it and state which of them are independent of temperature.

Answer: The concentration of a solution tells us the amount of solute dissolved in a given amount of solvent or solution. A dilute solution has little solute relative to the solvent, and a concentrated solution has a lot.
Ways of expressing concentration: mass percentage, volume percentage, mass by volume percentage, parts per million (ppm), mole fraction, molarity and molality.
Independent of temperature: mass percentage, mole fraction, molality and ppm, because they are all based on mass or number of moles, which do not change with temperature.
Depends on temperature: molarity and the volume-based percentages, because volume expands or contracts when the temperature changes.

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