Class 12 Chemistry Important Chapter 2 Solutions

Class 12 Chemistry Important Chapter 2 Solutions Solutions English Medium As Per The New Syllabus to each chapter is provided in the list so that you can easily browse through different chapters ASSEB Class 12 Chemistry Important Solutions in English and select need one. AHSEC Class 12 Chemistry Additional Notes Download PDF. HS 2nd Year Chemistry Additional Solutions.

Class 12 Chemistry Important Chapter 2 Solutions

Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. ASSEB Class 12 Chemistry Additional Question Answer are part of All Subject Solutions. Here we have given HS 2nd Year Chemistry Important Solutions English Medium for All Chapters, You can practice these here.

Solutions

Chapter: 2

PART – I

IMPORTANT QUESTION AND ANSWER

1. Define molality and give its SI unit.

Ans: Molality (m) is the number of moles of solute present in 1 kg of solvent.

2. What type of deviation from Raoult’s Law is shown by a mixture of ethanol and acetone? Why?

Ans: Positive deviation: Because hydrogen bonding in ethanol is disrupted by acetone, reducing A-B attraction and increasing vapour pressure.

3. Define azeotrope. Give one example.

Ans: Azeotrope is a binary mixture that boils at a constant temperature and has the same composition in liquid and vapour phase.

Example: Ethanol + water (95.6% ethanol) – minimum boiling azeotrope.

4. What is meant by abnormal molar mass?

Ans: A molar mass which deviates from the expected value due to association or dissociation of solute particles in solution is called abnormal molar mass.

5. Give two differences between ideal and non-ideal solutions.

Ans: (i) Ideal solution obeys Raoult’s Law at all concentrations; non-ideal does not.

(ii) ΔH_mix = 0 for ideal, ≠ 0 for non-ideal.

6. Define Henry’s Law constant. What does a high value of KH signify?

Ans: Henry’s Law constant (KH) relates gas pressure and mole fraction in solution:

p = KH⋅x

A high KH value implies low solubility of the gas in liquid.

7. Define colligative properties. Name any two such properties.

Ans: Colligative properties are those which depend only on the number of solute particles, not their nature.

Examples: (i) Boiling point elevation, (ii) Osmotic pressure.

8. Give relation between elevation in boiling point and molar mass of solute.

Ans:

Where ΔT_b = elevation in boiling point, w₂ = mass of solute, w₁ = mass of solvent, M₂ 

= molar mass of solute, Kb = boiling point constant.

9. What is reverse osmosis? Name one use.

Ans: When pressure higher than osmotic pressure is applied on the solution side, pure solvent flows out through a semipermeable membrane – this is reverse osmosis.

Use: Desalination of seawater.

10. What is the effect of temperature on solubility of gases in liquids?

Ans: The solubility of gases in liquids is inversely related to temperature. As the temperature increases, the solubility of most gases in liquids decreases. 

This behavior can be explained with the help of the following points:

(i) Nature of Dissolution Process: The dissolution of gases in liquids is generally an exothermic process, meaning it releases heat. When a gas dissolves in a liquid, gas molecules lose energy and become part of the liquid phase, which is a lower-energy state.

(ii) Le Chatelier’s Principle: According to Le Chatelier’s Principle, if a system at equilibrium experiences a change in temperature, the system adjusts to counteract that change. When temperature is increased, the system shifts in the direction that absorbs heat. Since dissolution of gases releases heat (exothermic), increasing temperature shifts the equilibrium in the reverse direction, thus decreasing solubility.

(iii) Kinetic Explanation: At higher temperatures, gas molecules gain kinetic energy. This increases their tendency to escape from the liquid back into the gaseous phase, thereby reducing the concentration of dissolved gas.

(iv) Practical Observations: In cold water, gases like oxygen and carbon dioxide dissolve more easily. That is why aquatic animals thrive better in cold water. Soft drinks lose their fizz faster when left in warm conditions because CO₂ escapes more rapidly due to reduced solubility at higher temperatures.

11. (i) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte? Justify your answer.

Ans: “B’ is a strong electrolyte.

Reason: A strong electrolyte is already dissociated into ions, but on dilution interionic forces are overcome, ions are free to move. Therefore, there is alight increase in molar conductivity on dilution.