Question

The number of atoms in $67.2L$ of $N{H_3}$ (g) at STP is:
A. $9{N_A}$
B. $12{N_A}$
C. $3{N_A}$
D. $4{N_A}$

Answer 1

In Chemistry, STP is "Standard temperature and pressure". Under these conditions, a gas occupies a volume of $22.4L$ and will have $6.023 \times {10^{23}}$ molecules. The number $6.023 \times {10^{23}}$ is known as Avogadro's number . Avogadro's number, denoted by ${N_A},$ is the number of constituent particles in one mole of any substance.

Complete answer:
In accordance with Avogadro's law, we know that one mole of a gas at STP or standard temperature and pressure has $6.023 \times {10^{23}}$ molecules, and occupies a volume of $22.4L.$
$\Rightarrow$ One mole of $N{H_3}$ gas at STP occupies $22.4L$ volume, and has $6.023 \times {10^{23}}$ or ${N_A}$ molecules.
We need to calculate the number of atoms in $67.2L$ of $N{H_3}$ gas at STP.
Since we know that
$22.4L = {N_A} \to (1)$
Let us take "x", which is the number of molecules in $67.2L$ of $N{H_3}$ gas. Then,
$67.2L = x \to (2)$
Therefore, by multiplying ${\text{(1) and (2)}}$ and isolating x to LHS, we get:
$x = \dfrac{{67.2}}{{22.4}} \times {N_A}$
$= 3{N_A}$
Here, we have obtained the number of molecules in $67.2L$ of $N{H_3}$ gas. Our aim is to find the number of atoms, and not the number of molecules. Considering that, we know that one molecule of $N{H_3}$ consists of one nitrogen atom and three hydrogen atoms. The total number of atoms in one molecule of $N{H_3}$ is four atoms. Using that information, let us calculate the required answer:
The number of atoms in $67.2L$ of $N{H_3}$ (g) at STP
$= 3{N_A} \times 4$
$= 12{N_A}$
Therefore, the required answer is B. $12{N_A}.$

Note:
While applying Avogadro's law, we need to be aware of what species that we are applying it to: whether it is an atom, or a molecule. This confusion might arise due to the fact that the units considered in Avogadro's law may be electrons, atoms, ions or molecules. This usually depends on the nature of the substance.