Match List-I with List-II and select the corr | vedclass

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(A) Step $1$: $(a)$ $2 	ext{ moles of ethene } (C_2H_4) = 2 	imes 28 	ext{ g} = 56 	ext{ g}$. Matches $(ii)$.Step $2$: $(b)$ Molar mass of $CO_2 =

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$a-ii, b-iv, c-i, d-iii$

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(A) Step $1$: $(a)$ $2 	ext{ moles of ethene } (C_2H_4) = 2 	imes 28 	ext{ g} = 56 	ext{ g}$. Matches $(ii)$.Step $2$: $(b)$ Molar mass of $CO_2 = 44 	ext{ g/mol}$. $1.5 	ext{ moles} = 1.5 	imes 44 	ext{ g} = 66 	ext{ g}$. Matches $(iv)$.Step $3$: $(c)$ $1 	ext{ g of } H_2 = \frac{1}{2} 	ext{ mole} = 0.5 	imes 6.022 	imes 10^{23} 	ext{ molecules} = 3.011 	imes 10^{23} 	ext{ molecules}$. Note: The provided options suggest a mismatch in the original question's $(c)$ and $(d)$ mapping. Based on standard calculations, $(c)$ corresponds to $0.5 	ext{ mole}$ and $(d)$ $2 	ext{ moles of } H_2O$ at $STP$ is $44.8 	ext{ L}$. Given the options, $(a-ii, b-iv, c-i, d-iii)$ is the intended logical set.

List-$I$	List-$II$
$(a)$ $2 \text{ moles of ethene}$	$(i)$ $11.2 \text{ L volume at STP}$
$(b)$ $\text{Molar mass is equal to } 66 \text{ g}$	$(ii)$ $56 \text{ g}$
$(c)$ $1 \text{ g of } H_2$	$(iii)$ $12.04 \times 10^{23} \text{ molecules}$
$(d)$ $2 \text{ moles of water vapours}$	$(iv)$ $1.5 \text{ mole of } CO_2$

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