CBSE Class 12th Chemistry Chapter 1 - Solutions Important Questions with Answers

You should focus on solving CBSE Class 12th Chemistry Chapter 1: Solutions important questions, especially to help you score high marks. By solving CBSE Class 12th Chemistry 1 questions, you will be solving exam-oriented questions only.

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CBSE Class 12th Chemistry Chapter 1 - Solutions: In the CBSE 12th Chemistry Chapter 1 - Solutions, students will get detailed information about the types of solutions, solubility of gases in liquids, expression of concentration of solutions of solids in liquids, solid solutions, colligative properties – relative lowering of vapour pressure, depression of freezing point, elevation of B.P., osmotic pressure, determination of molecular masses using colligative properties, and abnormal molecular mass.

The students should possess detailed knowledge of the solution to clarify complex concepts and improve problem-solving skills. Through the chapter, students can understand the concentration of solutions, the vapour pressure of liquid solutions, types of solutions, solubility of gases, solids in a liquid, ideal and non ideal solutions, and Raoult’s Law. The students will also get various problems based on finding the molarity, mass percentage, mole fraction, Henry’s Law, and more.

After completing the syllabus, students can practice the questions. They can check the questions given below while preparing for the CBSE Class 12th Chemistry Chapter 1 - Solutions.

Q.1 - Which type of solid solution will result by mixing two solid components with large differences in the sizes of their molecules?

Q.2 - Gases tend to be less soluble in liquids as the temperature is raised. Why?

Q.3 - A mixture of chlorobenzene and bromobenzene forms a nearly ideal solution but a mixture of chloroform and acetone does not. Why?

Q.4 - How relative lowering in vapor pressure is related with depression in freezing point and elevation in boiling point?

Q.5 - Why do colligative properties of solution of a given concentration are found to give abnormal molecular weight of solute. Explain with the help of suitable examples.

Q.6 - Two liquids X and Y on mixing form an ideal solution. The vapor pressure of the solution containing 2 mol of X and 1 mol of Y is 550 mm Hg. But when 4 mol of X and 1 mole of Y are mixed, the vapor pressure of solution thus formed is 560 mm Hg. What will be the vapor pressure of pure X and pure Y at this temperature?

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Define the following terms: (Delhi 2017)
(i) Colligative properties
(ii) Molality (m)

Question 2.

Define the following terms:
(i) Abnormal molar mass
(ii) van’t Hoff factor (i) (Delhi 2017)

Question 3.

Define the following terms:
(i) Ideal solution
(ii) Molarity (M) (Delhi 2017)

Question 4.

Explain why on addition of 1 mol of glucose to 1 litre of water, the boiling point of water increases. (Comptt. Delhi 2017)

Question 5.

A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol^-1) (Molar mass of glucose = 180 g mol^-1) (Delhi 2017)

Question 6.

The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mmHg respectively. Find out the composition of liquid mixture if total vapour pressure at this temperature is 600 mmHg. (Comptt. Delhi 2017)

Question 7.

(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol^-1) (Molar mass of glucose = 180 g mol^-1)
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass (All India 2017)

Question 8.

(a) 30 g of urea (M = 60 g mol^-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions. (All India 2017)

Question 9.

(a) Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases.
(b) Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the number of moles of CO₂ in 500 ml of soda water when packed under 2.53 × 10⁵ Pa at the same temperature. (Comptt. All India 2017)

Question 10.

(a) Define the following terms :
(i) Ideal solution (ii) Osmotic pressure
(b) Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl₂ to 200 g of water, assuming that CaCl₂ is completely dissociated.
(K_b) for water = 0.512 K kg mol^-1; Molar mass of CaCl₂ = 111 g mol^-1) (Comptt. All India 2017)

Question 1.

(i) Gas (A) is more soluble in water than Gas
(B) at the same temperature. Which one of the two gases will have the higher value of K_H (Henry’s constant) and why?
(ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes? (All India 2016)

Question 2.

What is osmotic pressure? Why it is a colligative property? (Comptt Delhi 2016)

Question 3.

Define osmotic pressure. How is osmotic pressure related to the concentration of a solute in a solution? (Comptt. All India 2016)

Question 4.

Calculate the boiling point of solution when 4 g of MgSO₄ (M =120 g mol^-1) was dissolved in 100 g of water, assuming MgSO₄ undergoes complete ionization.
(K_b for water = 0.52 K kg mol^-1) (All India 2016)

Question 5.

Calculate the mass of a non-volatile solute (molecular mass 40) which should be dissolved in 114 g octane to reduce the vapour pressure to 80%. (Comptt. Delhi 2016)

Question 6.

An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?
[Vapour pressure of water = 1.013 bar] (Comptt. All India 2016)

Question 7.

Calculate the freezing point of solution when 1.9 g of MgCl₂ (M = 95 g mol^-1) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionization.
(K_f for water = 1.86 K kg mol^-1)
(b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution? (Delhi 2016)

Question 8.

(a) When 2.56 g of sulphur was dissolved in 100 g of CS₂, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_x).
(K_f for CS₂ = 3.83 K kg mol^-1, Atomic mass of Sulphur = 32 g mol^-1)
(b) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?(Delhi 2016)