Animated Solution for Chemistry - Solutions: The vapour pressure of water at $20^{\circ }\text{C}$ is $17.5\text{ mmHg}$. If $18\text{ g}$ of glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) is added to $178.2\text{ g}$ of water at $20^{\circ }\text{C}$, the vapour pressure of the resulting solution will be

Imagine a closed container partially filled with pure water. At any given temperature, some water molecules possess enough kinetic energy to escape the liquid phase and enter the vapour phase.

These vapour molecules bounce around and exert a pressure on the walls of the container and the surface of the liquid. This is known as the vapour pressure of the pure solvent, denoted by $p^{\circ }$. In our case, at $20^{\circ }\text{C}$, this pressure is $17.5\text{ mmHg}$.

Now, we introduce a twist: we dissolve $18\text{ g}$ of glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) into $178.2\text{ g}$ of this water. Glucose is a non-volatile solute. This means it does not readily evaporate at this temperature.

When glucose molecules dissolve, they occupy space throughout the solution, including the surface. Because some of the surface area is now taken up by glucose molecules instead of water molecules, fewer water molecules can escape into the vapour phase. Consequently, the vapour pressure of the solution ($p_s$) drops. This phenomenon is called the lowering of vapour pressure.

To quantify this drop, we use Raoult's Law for solutions containing non-volatile solutes. It states that the relative lowering of vapour pressure is exactly equal to the mole fraction of the solute in the solution.

Mathematically, it is expressed as:

Before we can use our master equation, we need the mole fraction of glucose (${\chi }_{\text{glucose}}$). This requires finding the number of moles of both glucose and water.

1. Moles of Glucose: The molar mass of glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) is $180\text{ g/mol}$.

2. Moles of Water: The molar mass of water (${\text{H}}_2\text{O}$) is $18\text{ g/mol}$.

3. Mole Fraction of Glucose: The mole fraction is the ratio of the moles of glucose to the total moles in the solution.

Now, we substitute our known values back into Raoult's Law:

Multiplying both sides by $17.5$:

Finally, solving for the vapour pressure of the solution ($p_s$):

The vapour pressure has indeed lowered from $17.5\text{ mmHg}$ to $17.325\text{ mmHg}$. This elegant application of Raoult's Law perfectly predicts the physical behavior of the solution!