Question

Which has the maximum number of molecules?

• A. $2\mathrm{g}{\mathrm{H}}_2$
• B. $7\mathrm{g}{\mathrm{N}}_2$
• C. $16\mathrm{g}{\mathrm{O}}_2$
• D. $16\mathrm{g}{\mathrm{NO}}_2$

Answer 1

The correct option is A

$2\mathrm{g}{\mathrm{H}}_2$

Explanation for correct answer:-

Option (A) $2\mathrm{g}{\mathrm{H}}_2$

Step 1: Calculating the number of molecules

Molar mass of ${\mathrm{H}}_2$ is $1\times 2=2\mathrm{g}{\mathrm{mol}}^{-1}$

We know that, $\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{weight}}{\mathrm{molecular}\mathrm{weight}}$

$$\begin{gathered} \therefore \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{given}\mathrm{weight}\mathrm{of}{\mathrm{H}}_2}{\mathrm{molecular}\mathrm{weight}\mathrm{of}{\mathrm{H}}_2} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{2}{2} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=1 \end{gathered}$$

And we also that, 1 mole of any substance contains $6.022\times {10}^{23}$molecules (Avagadro's number)

Hence it is the correct option

Explanation for incorrect answer:-

Option (B) $7\mathrm{g}{\mathrm{N}}_2$

Step 1: Calculating the number of molecules

Molar mass of ${\mathrm{N}}_2$ is $2\times 14=28\mathrm{g}{\mathrm{mol}}^{-1}$

We know that, $\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{weight}}{\mathrm{molecular}\mathrm{weight}}$

$$\begin{gathered} \therefore \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{given}\mathrm{weight}\mathrm{of}{\mathrm{N}}_2}{\mathrm{molecular}\mathrm{weight}\mathrm{of}{\mathrm{N}}_2} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{7}{28} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=0.25 \end{gathered}$$

And we also that, 1 mole of any substance contains $6.022\times {10}^{23}$molecules (Avagadro's number)

$\therefore$0.25 moles will contain $6.022\times {10}^{23}\times 0.25=1.505\times {10}^{23}$molecules.

Hence it is the incorrect option

Option (C) $16\mathrm{g}{\mathrm{O}}_2$

Step 1: Calculating the number of molecules

Molar mass of $O_2$ is $2\times 16=32\mathrm{g}{\mathrm{mol}}^{-1}$

We know that, $\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{weight}}{\mathrm{molecular}\mathrm{weight}}$

$$\begin{gathered} \therefore \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{given}\mathrm{weight}\mathrm{of}{\mathrm{O}}_2}{\mathrm{molecular}\mathrm{weight}\mathrm{of}{\mathrm{O}}_2} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{16}{32} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=0.5 \end{gathered}$$

And we also that, 1 mole of any substance contains $6.022\times {10}^{23}$molecules (Avagadro's number)

$\therefore$0.5 moles will contain $6.022\times {10}^{23}\times 0.5=3.011\times {10}^{23}$molecules.

Hence it is the incorrect option

Option (D) $16\mathrm{g}{\mathrm{NO}}_2$

Step 1: Calculating the number of molecules

Molar mass of $N{O}_2$ is $2\times 16+14=46\mathrm{g}{\mathrm{mol}}^{-1}$

We know that, $\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{weight}}{\mathrm{molecular}\mathrm{weight}}$

$$\begin{gathered} \therefore \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{given}\mathrm{weight}\mathrm{of}{\mathrm{NO}}_2}{\mathrm{molecular}\mathrm{weight}\mathrm{of}{\mathrm{NO}}_2} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=\frac{16}{46} \\ \Rightarrow \mathrm{number}\mathrm{of}\mathrm{moles}=0.347 \end{gathered}$$

And we also that, 1 mole of any substance contains $6.022\times {10}^{23}$molecules (Avagadro's number)

$\therefore$0.33 moles will contain $6.022\times {10}^{23}\times 0.347=2.09\times {10}^{23}$molecules.

Hence it is the incorrect option.

Therefore, option (A) $2\mathrm{g}{\mathrm{H}}_2$ is the correct answer.