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NEET CHEMISTRYMedium

Match List-I with List-II:

| List-I (Quantities) | List-II (Corresponding Values) | | :--- | :--- | | (a) 4.48 litres of O2O_2 at STP | (i) 0.2 mole | | (b) 12.022×102212.022 \times 10^{22} molecules of H2OH_2O | (ii) 12.044×102312.044 \times 10^{23} molecules | | (c) 96 g of O2O_2 | (iii) 6.4 g | | (d) 88 g of CO2CO_2 | (iv) 67.2 litres at STP |

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

A

(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

B

(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

C

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

D

(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)

Step-by-Step Solution

  1. Analyze (a): 4.48 L O24.48 \text{ L } O_2 at STP. Moles=Volume22.4=4.4822.4=0.2 mol\text{Moles} = \frac{\text{Volume}}{22.4} = \frac{4.48}{22.4} = 0.2 \text{ mol} Matches with (i).

  2. Analyze (c): 96 g O296 \text{ g } O_2. Molar mass of O2=32 g/molO_2 = 32 \text{ g/mol}. Moles=9632=3 mol\text{Moles} = \frac{96}{32} = 3 \text{ mol} Volume at STP=3×22.4=67.2 L\text{Volume at STP} = 3 \times 22.4 = 67.2 \text{ L} Matches with (iv).

  3. Analyze (d): 88 g CO288 \text{ g } CO_2. Molar mass of CO2=44 g/molCO_2 = 44 \text{ g/mol}. Moles=8844=2 mol\text{Moles} = \frac{88}{44} = 2 \text{ mol} Molecules=2×6.022×1023=12.044×1023\text{Molecules} = 2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23} Matches with (ii).

  4. Analyze (b): 12.022×102212.022 \times 10^{22} molecules. Moles=12.022×10226.022×10230.2 mol\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 mol×32 g/mol (O2)=6.4 g0.2 \text{ mol} \times 32 \text{ g/mol } (O_2) = 6.4 \text{ g}. There is a likely typo in the question text specifying H2OH_2O instead of O2O_2, as 0.2 mol H2O0.2 \text{ mol } H_2O would be 3.6 g3.6 \text{ 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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