Animated Solution for Chemistry - States of Matter: If $10^{-4}{\text{ dm}}^3$ of water is introduced into a $1.0{\text{ dm}}^3$ flask at $300\text{ K}$, then how many moles of water are in the vapour phase when equilibrium is established? (Given, vapour pressure of ${\text{H}}_2\text{O}$ at $300\text{ K}$ is $3170\text{ Pa}$; $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$)

Imagine a closed flask with a total volume of $1.0{\text{ dm}}^3$. We introduce a tiny drop of water into it—specifically, $10^{-4}{\text{ dm}}^3$. Because the flask is closed and initially empty (or contains non-reacting air which doesn't affect the partial pressure of water), the water will begin to evaporate.

This evaporation won't continue forever. It stops when the space inside the flask becomes saturated with water vapour. At this point, a dynamic equilibrium is established between the liquid water and the water vapour. The pressure exerted by the water vapour at this equilibrium state is exactly its vapour pressure, which is given as $3170\text{ Pa}$ at $300\text{ K}$.

To find the number of moles of water in the vapour phase, we can treat the water vapour as an ideal gas. The ideal gas equation is our master tool here:

Before we plug in the numbers, we need to be extremely careful with our units. The pressure $p$ is already in standard SI units ($3170\text{ Pa}$). The temperature $T$ is in Kelvin ($300\text{ K}$). The universal gas constant $R$ is $8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

What about the volume $V$? The total volume of the flask is $1.0{\text{ dm}}^3$. The volume occupied by the liquid water is $10^{-4}{\text{ dm}}^3$. The volume available for the vapour is the difference between the two:

Since $10^{-4}$ is $0.0001$, the available volume is $0.9999{\text{ dm}}^3$. For all practical purposes, we can approximate this to $1.0{\text{ dm}}^3$. Now, we must convert this to the SI unit of volume, which is cubic meters (${\text{m}}^3$). Since $1\text{ dm}=10^{-1}\text{ m}$, it follows that $1{\text{ dm}}^3=10^{-3}{\text{ m}}^3$.

Now, we substitute these pristine values into our rearranged ideal gas equation:

Let's simplify the numerator and denominator:

Dividing these gives us the final number of moles:

In scientific notation, this is beautifully written as:

This perfectly matches option (d). It is a brilliant reminder that in physical chemistry, keeping a sharp eye on units and making valid approximations (like ignoring the liquid's volume) is the key to arriving at the correct answer swiftly.