Animated Solution for Chemistry - Solutions: At room temperature, a dilute solution of urea is prepared by dissolving $0.60\text{ g}$ of urea in $360\text{ g}$ of water. If the vapour pressure of pure water at this temperature is $35\text{ mm Hg}$, lowering of vapour pressure will be (Molar mass of urea $=60{\text{ g mol}}^{-1}$)

Vapour Pressure

Imagine a closed container partially filled with pure water. Even at room temperature, some water molecules possess enough kinetic energy to escape the liquid phase and enter the space above as a gas. These gaseous molecules bounce around, colliding with the walls of the container and the surface of the liquid, creating what we call vapour pressure.

Now, what happens when we introduce a non-volatile guest into this environment? In our problem, we dissolve urea into the water. Urea is a non-volatile solute, meaning it has absolutely no desire to vaporize. As these urea molecules mix into the water, they occupy precious real estate at the surface of the liquid. With fewer water molecules at the surface, the rate at which water can escape into the vapour phase drops. Consequently, the overall vapour pressure of the solution decreases. This phenomenon is known as the lowering of vapour pressure.

Raoult's Law

To quantify exactly how much the pressure drops, we turn to the elegant mathematics of Raoult's Law. For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure is directly proportional to the mole fraction of the solute.

Mathematically, it is expressed as:

Here, $\Delta p$ is the lowering of vapour pressure, $p^{\circ }$ is the vapour pressure of the pure solvent, ${\chi }_B$ is the mole fraction of the solute, and $i$ is the van't Hoff factor. Our goal is to isolate and calculate $\Delta p$.

Moles and Fractions

Before we can unlock the final answer, we need to gather our variables. Let's start by calculating the number of moles of our solute (urea) and solvent (water).

For urea, we are given a mass of $0.60\text{ g}$ and a molar mass of $60{\text{ g mol}}^{-1}$:

For water, we have a massive $360\text{ g}$ pool. With a molar mass of $18{\text{ g mol}}^{-1}$:

Notice the stark contrast! We have $20\text{ moles}$ of solvent and merely $0.01\text{ moles}$ of solute. This confirms we are dealing with a highly dilute solution. Next, we calculate the mole fraction of urea (${\chi }_B$):

Calculating the Drop

We are almost there. We need one final piece: the van't Hoff factor ($i$). Because urea is a covalent, non-electrolyte compound, it does not dissociate into ions when dissolved in water. Therefore, one molecule of urea remains exactly one particle in solution, giving us $i=1$.

Now, we substitute all our hard-earned values back into Raoult's Law to find the absolute drop in pressure:

Rounding to the nearest significant decimal provided in the options, we arrive at $0.017\text{ mm Hg}$. The presence of just a tiny amount of urea was enough to measurably suppress the escaping tendency of the water molecules. Physics in action!