Chapter Exam

Unit 2: Solutions — Online MCQ Test

Q: What is the molarity of a solution containing 0.5 mol of NaOH in 1.0 L of solution?

Answer: A. 0.5 M — Molarity is defined as moles of solute per litre of solution. Here, 0.5 mol in 1.0 L gives 0.5 M.

Q: The vapour pressure of a solvent in a solution containing a non-volatile solute is 90 mm Hg, while that of the pure solvent is 100 mm Hg. The relative lowering of vapour pressure is:

Answer: B. 0.10 — Relative lowering of vapour pressure is (p0 - p)/p0 = (100 - 90)/100 = 0.10. This is a colligative property.

CHEMISTRY · CLASS 12 INTERMEDIATE 2 YEAR · Telangana State Board

Practice Unit 2: Solutions with a free chapter-wise online MCQ test for Telangana State Board CLASS 12 INTERMEDIATE 2 YEAR CHEMISTRY. This chapter covers: This chapter details solution types concentration units Henry Law Raoult Law ideal non-ideal solutions and colligative properties including osmotic pressure and abnormal molar mass.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 2: Solutions — Important Questions & Answers (FAQ)

Frequently asked questions from Telangana State Board CLASS 12 INTERMEDIATE 2 YEAR CHEMISTRY — Unit 2: Solutions, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is the molarity of a solution containing 0.5 mol of NaOH in 1.0 L of solution?

A. 0.5 M ✓
B. 1.0 M
C. 2.0 M
D. 0.05 M

Answer: A. 0.5 M
Molarity is defined as moles of solute per litre of solution. Here, 0.5 mol in 1.0 L gives 0.5 M.

The mole fraction of a component in a binary solution is defined as:

A. Moles of the component divided by total moles of all components ✓
B. Mass of the component divided by total mass of solution
C. Moles of the component divided by volume of solution
D. Mass of solute divided by mass of solvent

Answer: A. Moles of the component divided by total moles of all components
Mole fraction is the ratio of the number of moles of one component to the total number of moles present in the solution.

A solution shows negative deviation from Raoult’s law when:

A. A–B interactions are weaker than A–A and B–B interactions
B. A–B interactions are stronger than A–A and B–B interactions ✓
C. The solution is ideal
D. The vapor pressure becomes zero

Answer: B. A–B interactions are stronger than A–A and B–B interactions
Negative deviation occurs when unlike molecule interactions are stronger, causing lower vapour pressure than expected from Raoult’s law.

A solute in solution shows association. Which of the following is most likely observed in the determination of molar mass by colligative property methods?

A. Observed molar mass is lower than actual molar mass
B. Observed molar mass is higher than actual molar mass ✓
C. Observed molar mass is unchanged
D. Observed molar mass becomes zero

Answer: B. Observed molar mass is higher than actual molar mass
Association decreases the number of particles in solution, causing the colligative property to be smaller and the calculated molar mass to appear higher.

The vapour pressure of a solvent in a solution containing a non-volatile solute is 90 mm Hg, while that of the pure solvent is 100 mm Hg. The relative lowering of vapour pressure is: