Case Study Questions for Class 12 Chemistry Chapter 2 Solutions

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Board	CBSE
Useful for	Class 12 Students
Subject	Chemistry
Chapter	Chapter 2 Solutions
Type of Questions	Case Study
Answers Provided	Yes
Format	Question-Answer Format
Important Link	Class 12 Chemistry Case Study Questions

Table of Contents

Case Study Questions for Class 12 Chemistry Chapter 2 Solutions

Class 12 Chemistry · Chapter 2

There is Case Study Questions in class 12 Chemistry in session 2020-21. For the first time, the board has introduced the case study questions in the board exam. The first two questions in the board exam question paper will be based on Case Study and Assertion & Reason. The first question will have 5 MCQs out of which students will have to attempt any 4 questions. The second question will carry 5 Assertion & Reason type questions with the choice to attempt any four. Here are the questions based on case study.

Case Study 1: Colligative Properties and Vapour Pressure

Read the passage carefully, then answer the questions

Ch 2 · Solutions

The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties.

For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg. (Vapour pressure of water at $20^\circ\text{C}$ is 17.5 mm of Hg)

Relative lowering of vapour pressure for the given solution is

a 0.00348

b 0.061

c 0.122

d 1.75

Correct Answer (a) 0.00348 ✅

Explanation

The relative lowering of vapour pressure is the ratio of the lowering in vapour pressure ($\Delta P$) to the vapour pressure of the pure solvent ($P^\circ$).

$$\text{Relative Lowering} = \frac{\Delta P}{P^\circ}$$

$$= \frac{0.061}{17.5} \approx 0.00348$$

The vapour pressure (mm of Hg) of solution will be

a 17.5

b 0.61

c 17.439

d 0.00348

Correct Answer (c) 17.439 ✅

Explanation

The vapour pressure of the solution ($P_s$) is the vapour pressure of the pure solvent ($P^\circ$) minus the lowering in vapour pressure ($\Delta P$).

$$P_s = P^\circ – \Delta P$$

$$P_s = 17.5 – 0.061 = 17.439\text{ mm of Hg}$$

Mole fraction of sugar in the solution is

a 0.00348

b 0.9965

c 0.061

d 1.75

Correct Answer (a) 0.00348 ✅

Explanation

According to Raoult’s Law, the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute ($X_{\text{solute}}$).

$$X_{\text{sugar}} = \frac{\Delta P}{P^\circ} = \frac{0.061}{17.5} = 0.00348$$

If weight of sugar taken is 5 g in 108 g of water then molar mass of sugar will be

a 358

b 120

c 240

d 400

Correct Answer (c) 240 ✅

Explanation

For a dilute solution, the mole fraction of the solute $X_B \approx \frac{n}{N}$, where $n$ is moles of solute and $N$ is moles of solvent.

$$n = \frac{5}{M} \quad \text{and} \quad N = \frac{108}{18} = 6\text{ moles}$$

$$X_B = \frac{n}{N} \implies 0.00348 = \frac{5/M}{6}$$

$$M = \frac{5}{6 \times 0.00348} = \frac{5}{0.02088} \approx 239.46\text{ g/mol}$$

The closest value provided in the options is 240.

The vapour pressure (mm of Hg) of water at 293K when 25g of glucose is dissolved in 450 g of water is

a 17.2

b 17.4

c 17.120

d 17.02

Correct Answer (b) 17.4 ✅

Explanation

Molar mass of glucose ($C_6H_{12}O_6$) $= 180\text{ g/mol}$.
Moles of glucose ($n$) $= \frac{25}{180} = 0.1388\text{ mol}$.
Moles of water ($N$) $= \frac{450}{18} = 25\text{ mol}$.

Mole fraction of glucose ($X_B$):

$$X_B = \frac{n}{n+N} = \frac{0.1388}{25.1388} \approx 0.00552$$

Using relative lowering of vapour pressure ($\Delta P / P^\circ = X_B$):

$$\frac{17.5 – P_s}{17.5} = 0.00552$$

$$17.5 – P_s = 17.5 \times 0.00552 = 0.0966$$

$$P_s = 17.5 – 0.0966 = 17.403\text{ mm of Hg}$$

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