Skip to main content
Chapter 1 of 5
Flashcards

Solutions

Punjab Board · Class 12 · Chemistry

Flashcards for Solutions — Punjab Board Class 12 Chemistry. Quick Q&A cards covering key concepts, definitions, and formulas.

45 questions36 flashcards5 concepts

Interactive on Super Tutor

Studying Solutions? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for flashcards and more.

1,000+ Class 12 students started this chapter today

A graph illustrating Raoult's Law for an ideal binary liquid-liquid solution, showing the partial vapor pressures of components and total vapor pressure as a function of mole fraction.
Super Tutor

Super Tutor has 5+ illustrations like this for Solutions alone — flashcards, concept maps, and step-by-step visuals.

See them all
36 Flashcards
Card 1Types of Solutions

Define a solution and explain what homogeneous mixture means.

Answer

A solution is a homogeneous mixture of two or more components that has uniform composition and properties throughout. The component present in the largest quantity is the solvent, and components prese

Card 2Types of Solutions

What are the six main types of solutions based on the physical states of solute and solvent?

Answer

The six types are: 1) Gas in gas (N₂ and O₂ mixture), 2) Gas in liquid (O₂ in water), 3) Liquid in liquid (ethanol in water), 4) Solid in liquid (glucose in water), 5) Liquid in solid (mercury in sodi

Card 3Expressing Concentration

Calculate the mass percentage of glucose if 20g of glucose is dissolved in 80g of water.

Answer

Mass % = (mass of solute / total mass of solution) × 100. Step 1: Total mass = 20g + 80g = 100g. Step 2: Mass % of glucose = (20g / 100g) × 100 = 20%. Step 3: Mass % of water = (80g / 100g) × 100 = 80

Card 4Expressing Concentration

What is mole fraction? Calculate the mole fraction of ethanol in a solution containing 46g of ethanol (C₂H₆O) and 90g of water (H₂O).

Answer

Mole fraction (x) = moles of component / total moles of all components. Step 1: Molar mass of C₂H₆O = 46 g/mol, H₂O = 18 g/mol. Step 2: Moles of ethanol = 46/46 = 1 mol. Step 3: Moles of water = 90/18

Card 5Expressing Concentration

Calculate the molarity of a solution containing 5.85g of NaCl dissolved in 500 mL of solution. (Molar mass of NaCl = 58.5 g/mol)

Answer

Molarity (M) = moles of solute / volume of solution in liters. Step 1: Moles of NaCl = 5.85g / 58.5 g/mol = 0.1 mol. Step 2: Volume = 500 mL = 0.5 L. Step 3: Molarity = 0.1 mol / 0.5 L = 0.2 M. Answer

Card 6Expressing Concentration

What is molality? Calculate the molality of 2.5g of acetic acid (CH₃COOH) dissolved in 75g of benzene.

Answer

Molality (m) = moles of solute / mass of solvent in kilograms. Step 1: Molar mass of C₂H₄O₂ = 60 g/mol. Step 2: Moles of acetic acid = 2.5g / 60 g/mol = 0.0417 mol. Step 3: Mass of benzene = 75g / 100

Card 7Henry's Law and Gas Solubility

What does Henry's Law state? Write the mathematical expression and explain the constants.

Answer

Henry's Law: The partial pressure of a gas in vapor phase is proportional to the mole fraction of the gas in the solution. Mathematical expression: p = K_H × x. Where: p = partial pressure of gas, K_H

Card 8Henry's Law and Gas Solubility

N₂ gas is bubbled through water at 293K. If N₂ exerts a partial pressure of 0.987 bar and K_H for N₂ = 76.48 kbar, how many millimoles of N₂ dissolve in 1 litre of water?

Answer

Step 1: Apply Henry's Law: x(N₂) = p / K_H = 0.987 bar / 76,480 bar = 1.29 × 10⁻⁵. Step 2: For 1 litre of water, moles of water = 55.5 mol. Step 3: x(N₂) = n(N₂) / (n(N₂) + 55.5) ≈ n(N₂) / 55.5 (since

+28 more flashcards available

Practice All

Frequently Asked Questions

What are the important topics in Solutions for Punjab Board Class 12 Chemistry?
Key topics in Solutions include Classification of different types of solutions based on physical states of solute and solvent, Step-by-step process for calculating mass percentage of a solution, Overview of different concentration expression methods and their applications. These are the concepts Punjab Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Solutions — Punjab Board Class 12 Chemistry?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many flashcards are available for Solutions?
There are 36 flashcards for Solutions covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

For serious students

Get the full Solutions chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for Punjab Board Class 12 Chemistry.