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) $12.022 \times 10^{22}$ molecules of $H_2O$ | (ii) $12.044 \times 10^{23}$ 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)

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

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

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

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

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