Question

In which case is the number of molecules of water maximum?

• A. ${10}^{-3}$ mol of water
• B. $0.00224\mathrm{L}$of water vapour at $1\mathrm{atm}$ and $273\mathrm{k}$
• C. $0.18\mathrm{g}$ of water.
• D. $18\mathrm{ml}$ of water

Answer 1

The correct option is D

$18\mathrm{ml}$ of water

The explanation for the correct answer:

Option (D) $18\mathrm{ml}$ of water

Step 1 Mass of water:

The given volume of water is $18\mathrm{ml}$.

The density of water is $1\mathrm{gm}/\mathrm{ml}$.

Mass of the given water is,

$$\begin{gathered} \mathrm{Mass}=\mathrm{Volume}\times \mathrm{Density} \\ \mathrm{Mass}=18\times 1 \\ \mathrm{Mass}=18\mathrm{gm} \end{gathered}$$

Step 2 Number of moles:

The molar mass of water is$18\mathrm{gm}$

The number of moles in water is

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}} \\ \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{18}{18} \\ \mathrm{Number}\mathrm{of}\mathrm{moles}=1 \end{gathered}$$

Step 3 Number of molecules:

The number of molecules of water is,

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{molecules}=\mathrm{Number}\mathrm{of}\mathrm{moles}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}=1\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}=6.02\times {10}^{23} \end{gathered}$$

Hence correct option.

The explanation for Incorrect Answer:

Option (A) ${10}^{-3}$ mol of water

Step 1 Number of molecules:

The number of molecules of water is,

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{molecules}=\mathrm{Number}\mathrm{of}\mathrm{moles}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}={10}^{-3}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}=6.02\times {10}^{20} \end{gathered}$$

Hence incorrect option.

Option (B) $0.00224\mathrm{L}$of water vapour at $1\mathrm{atm}$ and $273\mathrm{k}$

Step 1 Number of moles.

According to the ideal gas equation,

$\mathrm{PV}=\mathrm{nRT}$…..(1)

Where $\mathrm{P}$ is the pressure, $\mathrm{V}$ is the volume, $\mathrm{n}$ is the number of moles, $\mathrm{T}$ is the temperature and $\mathrm{R}$ is the gas constant.

The given pressure $\mathrm{P}=1\mathrm{atm}$

The given volume $\mathrm{V}=0.00224\mathrm{L}$

The given temperature $\mathrm{T}=273\mathrm{K}$

Gas constant $\mathrm{R}=0.0821\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

By rearranging equation (1)

$\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}$

Substituting the values we get,

$$\begin{gathered} \mathrm{n}=\frac{1\times 0.00224}{0.0821\times 273} \\ n={10}^{-4}\mathrm{moles} \end{gathered}$$

Step 2 Number of molecules:

The number of molecules of water is,

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{molecules}=\mathrm{Number}\mathrm{of}\mathrm{moles}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}={10}^{-4}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}=6.02\times {10}^{19} \end{gathered}$$

Hence incorrect option.

Option (C) $0.18\mathrm{g}$ of water:

Step 1 Number of moles:

The molar mass of water is$0.18\mathrm{gm}$

The number of moles in water is

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}} \\ \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{0.18}{18} \\ \mathrm{Number}\mathrm{of}\mathrm{moles}={10}^{-2} \end{gathered}$$

Step 2 Number of molecules:

The number of molecules of water is,

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{molecules}=\mathrm{Number}\mathrm{of}\mathrm{moles}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}={10}^{-2}\times 6.02\times {10}^{23} \\ \mathrm{Number}\mathrm{of}\mathrm{molecules}=6.02\times {10}^{21} \end{gathered}$$

Hence incorrect option.

Therefore, the number of molecules of water is maximum in Option (D) $18\mathrm{ml}$ of water.