Match List I with List II

	List - I		List - II ($\Delta_0$)
A.	16 g of $\mathrm{CH_4~(g)}$	I.	Weighs 28 g
B.	1 g of $\mathrm{H_2~(g)}$	II.	$60.2\times10^{23}$ electrons
C.	1 mole of $\mathrm{N_2~(g)}$	III.	Weighs 32 g
D.	0.5 mol of $\mathrm{SO_2~(g)}$	IV.	Occupies 11.4 L volume of STP

Choose the correct answer from the options given below:

<p>A. 16 g of $\mathrm{CH_4~(g)}$</p> <p>The molar mass of $\mathrm{CH_4}$ (Methane) is 16 g/mol. Therefore, 16 g of $\mathrm{CH_4}$ is equivalent to 1 mole of $\mathrm{CH_4}$. Furthermore, each molecule of $\mathrm{CH_4}$ has 10 electrons (6 from Carbon and 4 from Hydrogen). As a result, 1 mole of $\mathrm{CH_4}$ (or $6.022 \times 10^{23}$ molecules of $\mathrm{CH_4}$) would have $10 \times 6.022 \times 10^{23} = 60.2 \times 10^{23}$ electrons. This matches with (II).</p> <p>B. 1 g of $\mathrm{H_2~(g)}$</p> <p>The molar mass of $\mathrm{H_2}$ (Hydrogen) is 2 g/mol. Therefore, 1 g of $\mathrm{H_2}$ is equivalent to 0.5 moles of $\mathrm{H_2}$. The volume that a given quantity of gas occupies is proportional to the number of moles of gas. At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, 0.5 moles of gas would occupy $0.5 \times 22.4 = 11.2$ liters. The closest match is (IV) with 11.4 liters (the small discrepancy may be due to rounding or slightly different conditions than STP).</p> <p>C. 1 mole of $\mathrm{N_2~(g)}$</p> <p>The molar mass of $\mathrm{N_2}$ (Nitrogen) is 28 g/mol. Therefore, 1 mole of $\mathrm{N_2}$ weighs 28 g. This matches with (I).</p> <p>D. 0.5 mol of $\mathrm{SO_2~(g)}$</p> <p>The molar mass of $\mathrm{SO_2}$ (Sulfur Dioxide) is 64 g/mol. Therefore, 0.5 moles of $\mathrm{SO_2}$ weigh $0.5 \times 64 = 32$ g. This matches with (III).</p> <p>Thus, the correct matches are:</p> <p>A - II</p> <p>B - IV</p> <p>C - I</p> <p>D - III</p>