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Column – I and Column-II contain four entries in each. Entries of Column-I are to be matched with some entries of Column-II One or more than one entries of Column-I may have the matching with the same entries of Column-II
Column-I Column-II
$(A)$ 44 g $C O_{2}$ gas $(P)$ 1g molecule
$(B)$ 35.2 g of $C H_{4}$ $(Q)$ $N_{A}$ molecule
$(C)$ 48 g of $O_{3}$ gas $(R)$ $22 N_{A}$ electrons
$(D)$ 44 g of $N_{2} O$ gas $(S)$ 49.28 L at 1 atm and 273 k
$(T)$ $N_{A}$ atoms of oxygen

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$a \to p, q, r; b \to r, s; c \to p, q; d \to p, r$

$a \to p, q; b \to r, s; c \to p, q; d \to p, q, r, t$

$a \to p, q, r; b \to r, s; c \to p, s; d \to p, t$

$a \to p, q, r; b \to r, s; c \to p, q; d \to p, q, r, t$

answer is D.

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Detailed Solution

$(A) M o l e s o f C O_{2} (g) = \frac{44}{44} = 1$
$1 m o l e C O_{2} (g) = 1 g - m o l e c u l e C O_{2} (g)$
$(B) M o l e s o f C H_{4} (g) = \frac{35.2}{16} = 2.2 m o l e s$
$V = 2.2 \times 22.4 = 49.28 L a t 1 a t m a n d 273 k$
$(C) M o l e s o f O_{3} (g) = \frac{48}{48} = 1$
$(D) M o l e s o f N_{2} O (g) = \frac{44}{44} = 1$
$N u m b e r o f O a t o m s = N_{A} \times 1 = N_{A}$

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