JEE Chemistry 2023 Question with Solution

Given: Match List I with List II.

Find: The correct matching option.

For A, $16 \, \text{g of CH}_4\text{(g)} = 1$ mole of $\text{CH}_4$. One molecule of $\text{CH}_4$ has $10$ electrons, so $1$ mole contains

$$10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$$

electrons. Hence A-II.

For B, $1 \, \text{g of H}_2\text{(g)} = 0.5$ mole of $\text{H}_2$. At STP, $1$ mole occupies $22.7 \, \text{L}$, so $0.5$ mole occupies

$$\frac{22.7}{2} = 11.35 \, \text{L}$$

which corresponds to approximately $11.4 \, \text{L}$. Hence B-IV.

For C, $1$ mole of $\text{N}_2 = 28 \, \text{g}$. Hence C-I.

For D, $0.5$ mole of $\text{SO}_2$ has mass

$$0.5 \times 64 = 32 \, \text{g}$$

Hence D-III.

So the full matching is A-II, B-IV, C-I, D-III.

The solution concludes "The Correct Option is C", but the extracted working gives A-II, B-IV, C-I, D-III, which matches option A. Therefore, there is a discrepancy between the displayed option label and the working.

Given:

• A: $16 \, \text{g of CH}_4\text{(g)}$
• B: $1 \, \text{g of H}_2\text{(g)}$
• C: $1$ mole of $\text{N}_2\text{(g)}$
• D: $0.5$ mol of $\text{SO}_2\text{(g)}$

Find: The correct item in List II for each entry.

1. A: $16 \, \text{g of CH}_4$

$$\text{Molar mass of CH}_4 = 16 \, \text{g mol}^{-1}$$

So this is $1$ mole of $\text{CH}_4$. Each molecule has $6 + 4 = 10$ electrons. Thus total electrons in $1$ mole are

$$10 \times N_A = 10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$$

So A-II.

2. B: $1 \, \text{g of H}_2$

$$\text{Moles of H}_2 = \frac{1}{2} = 0.5$$

At STP,

$$1 \text{ mole gas} = 22.7 \, \text{L}$$

Therefore,

$$0.5 \text{ mole} = 11.35 \, \text{L} \approx 11.4 \, \text{L}$$

So B-IV.

3. C: $1$ mole of $\text{N}_2$

$$\text{Molar mass of N}_2 = 28 \, \text{g mol}^{-1}$$

So C-I.

4. D: $0.5$ mol of $\text{SO}_2$

$$\text{Molar mass of SO}_2 = 32 + 2 \times 16 = 64 \, \text{g mol}^{-1}$$

Hence mass is

$$0.5 \times 64 = 32 \, \text{g}$$

So D-III.

Therefore, the matching is A-II, B-IV, C-I, D-III.