Question

Which of the following contains a maximum number of molecules?

• A. 4 g of $O_2$
• B. 4 g of ${NH}_3$
• C. 4 g of ${CO}_2$
• D. 4 g of ${SO}_2$

Answer 1

The correct option is B

4 g of ${\mathrm{NH}}_3$

Total number of molecules is equal to the( number of moles $\times$ ${\mathrm{N}}_{\mathrm{A}}$ ( Avogadro's constant)).

Explanation for the correct options: (B) for 4 g of ${\mathrm{NH}}_3$

$$\begin{gathered} Numberofmoles=\frac{Weight}{MolecularWeight} \\ =\frac{4}{17} \\ Numberofmolcules=\frac{4}{17}\times N_A \\ =0.23N_A \end{gathered}$$

Explanation for the incorrect options: (A) 4 g of ${\mathrm{O}}_2$.

$numberofmoles=\frac{Weight}{MolecularWeight}$

$$\begin{gathered} =\frac{4}{32} \\ =\frac{1}{8} \end{gathered}$$

$$\begin{gathered} numberofmolecules=\frac{1}{8}\times N_A \\ =0.125N_A \end{gathered}$$

(C) 4 g of ${\mathrm{CO}}_2$.

$$\begin{gathered} Numberofmoles=\frac{Weight}{MolecularWeight} \\ =\frac{4}{28} \\ =\frac{1}{7} \\ Numberofmolcules=\frac{1}{7}{N}_A \\ =0.14N_A \end{gathered}$$

(D) 4 g of ${\mathrm{SO}}_2$

$$\begin{gathered} Numberofmoles=\frac{Weight}{MolecularWeight} \\ =\frac{4}{64} \\ =\frac{1}{16} \\ Numberofmolcules=\frac{1}{16}\times N_A \\ =\frac{1}{16}{N}_A \\ =0.06N_A \end{gathered}$$

Hence, option (B) is correct. 4 g of ${\mathrm{NH}}_3$ contains a maximum number of molecules.