Match List-I with List-II:

| List-I (Quantities) | List-II (Corresponding Values) |
| :--- | :--- |
| (a) 4.4

Question

Match List-I with List-II:

| List-I (Quantities) | List-II (Corresponding Values) |
| :--- | :--- |
| (a) 4.48 litres of $O_2$ at STP | (i) 0.2 mole |
| (b) $12.022 	imes 10^{22}$ molecules of $H_2O$ | (ii) $12.044 	imes 10^{23}$ molecules |
| (c) 96 g of $O_2$ | (iii) 6.4 g |
| (d) 88 g of $CO_2$ | (iv) 67.2 litres at STP |

(Given - Molar volume of a gas at STP = 22.4 L)

Accepted Answer

1. **Analyze (a):** $4.48 	ext{ L } O_2$ at STP.
   $$ 	ext{Moles} = \frac{	ext{Volume}}{22.4} = \frac{4.48}{22.4} = 0.2 	ext{ mol} $$
   Matches with **(i)**.

2. **Analyze (c):** $96 	ext{ g } O_2$.
   Molar mass of $O_2 = 32 	ext{ g/mol}$.
   $$ 	ext{Moles} = \frac{96}{32} = 3 	ext{ mol} $$
   $$ 	ext{Volume at STP} = 3 	imes 22.4 = 67.2 	ext{ L} $$
   Matches with **(iv)**.

3. **Analyze (d):** $88 	ext{ g } CO_2$.
   Molar mass of $CO_2 = 44 	ext{ g/mol}$.
   $$ 	ext{Moles} = \frac{88}{44} = 2 	ext{ mol} $$
   $$ 	ext{Molecules} = 2 	imes 6.022 	imes 10^{23} = 12.044 	imes 10^{23} $$
   Matches with **(ii)**.

4. **Analyze (b):** $12.022 	imes 10^{22}$ molecules.
   $$ 	ext{Moles} = \frac{12.022 	imes 10^{22}}{6.022 	imes 10^{23}} \approx 0.2 	ext{ mol} $$
   By elimination, this maps to **(iii) 6.4 g**. (Note: $0.2 	ext{ mol} 	imes 32 	ext{ g/mol } (O_2) = 6.4 	ext{ g}$. There is a likely typo in the question text specifying $H_2O$ instead of $O_2$, as $0.2 	ext{ mol } H_2O$ would be $3.6 	ext{ g}$, but the pairing logic with options confirms the match to iii).

5. **Conclusion:** The correct sequence is (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii).

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