Question

Which of the following contains the maximum number of atoms at STP?

• A. $44.8\text{L of}N{H}_3$
• B. $134.4\text{L of}{O}_2$
• C. $123.2\text{L of}S{O}_2$
• D. $44.8\text{L of}C{O}_2$

Answer 1

The correct option is C $123.2\text{L of}S{O}_2$

$\text{Moles of}N{H}_3=\frac{\text{given volume}}{\text{molar volume at STP}}=\frac{44.8}{22.4}=2\text{moles}$
$\text{Moles of}{O}_2=\frac{\text{given volume}}{\text{molar volume at STP}}=\frac{134.4}{22.4}=6\text{moles}$
$\text{Moles of}S{O}_2=\frac{\text{given volume}}{\text{molar volume at STP}}=\frac{123.2}{22.4}=5.5\text{moles}$
$\text{Moles of}C{O}_2=\frac{\text{given volume}}{\text{molar volume at STP}}=\frac{44.8}{22.4}=2\text{moles}$

1 mole of $N{H}_3$ contains 4 $N_A$ atoms
So, 2 moles of $N{H}_3$ will contain $(2\times 4)=8N_A$ atoms
1 mole of $O_2$ contains 2 $N_A$ atoms
So, 6 moles of $O_2$ will contain $(2\times 6)=12N_A$ atoms
1 mole of $S{O}_2$ contains 3 $N_A$ atoms
So, 5.5 moles of $S{O}_2$ will contain $(3\times 5.5)=16.5N_A$ atoms
1 mole of $C{O}_2$ contains 3 $N_A$ atoms
2 moles of $C{O}_2$ will contain $(2\times 3)=6N_A$ atoms