QJEEM23Chbe0b: JEE Main 2023 Chemistry - Class 11 JEEM23Chbe0b Solution: Match List I with List II .tg…

A. 16 g of $\mathrm{CH_4~(g)}$

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).

B. 1 g of $\mathrm{H_2~(g)}$

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).

C. 1 mole of $\mathrm{N_2~(g)}$

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).

D. 0.5 mol of $\mathrm{SO_2~(g)}$

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).

Thus, the correct matches are:

A - II

B - IV

C - I

D - III