Question

The number of atoms in 67.2 L of ${NH}_3$(g) at STP is:

• A. 9 $N_A$
• B. 12 $N_A$
• C. 3 $N_A$
• D. 4 $N_A$

Answer 1

Answer: (B)

12 $N_A$

By Avogadro law,

22.4 liter of any gas at STP$=$1mole

$=6.022\times 10^{23}$ molecules of gas

$=$NA (Avogadro number)

So, in 67.2 L of ${NH}_3=?$

22.4 L of ${NH}_3=N_A$

1 L of ${NH}_3=\frac{NA}{22.4}$

67.2 L of ${NH}_3=\frac{67.2}{22.4}\times NA$

$=3N_A$ molecules

Now, each molecule contain 4 atoma

Number of atoms $=4\times 3N_A=12N_A$ atoms