JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Chapter_pending

Convert each sample to moles and read off its property. 16 g of $\mathrm{C}{\mathrm{H}}_4$ is 1 mol, and each molecule has 10 electrons, so it holds $10\times 6.02\times 10^{23}=60.2\times 10^{23}$ electrons. 1 g of ${\mathrm{H}}_2$ is 0.5 mol, which at STP occupies $0.5\times 22.7=11.4$ L. 1 mol of ${\mathrm{N}}_2$ weighs 28 g. 0.5 mol of $\mathrm{S}{\mathrm{O}}_2$ weighs $0.5\times 64=32$ g. Hence $\mathrm{C}{\mathrm{H}}_4$ matches the $60.2\times 10^{23}$ electrons, ${\mathrm{H}}_2$ matches the 11.4 L volume, ${\mathrm{N}}_2$ matches 28 g and $\mathrm{S}{\mathrm{O}}_2$ matches 32 g.