Animated Solution for Chemistry - Solutions: 18 g of glucose ($C_6{H}_{12}{O}_6$) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is

How Glucose Lowers Vapour Pressure

Imagine a bustling microscopic city at the surface of a glass of pure water. Water molecules are constantly jumping out into the air (evaporating) and diving back in (condensing). When this chaotic dance reaches an equilibrium, the pressure exerted by the escaped water molecules is what we call the vapour pressure. At its normal boiling point, pure water pushes back against the atmosphere with a force of exactly $760\text{ torr}$.

But what happens when we introduce an intruder? In this problem, we add $18\text{ g}$ of glucose ($C_6{H}_{12}{O}_6$) to $178.2\text{ g}$ of water. Glucose is a non-volatile solute—it has no desire to evaporate. Instead, its bulky molecules sit lazily at the surface of the water, acting like microscopic bouncers. They physically block the water molecules from escaping, while doing nothing to stop them from returning. The result? Fewer water molecules in the air, and consequently, a lowered vapour pressure.

To figure out exactly how much the pressure drops, we need to know the ratio of our players. Chemistry operates in moles, not grams, so let's convert our masses.

First, the glucose:

Next, the water:

Now, we find the mole fraction of our non-volatile bouncer, glucose. This tells us what percentage of the total molecules are glucose.

So, exactly $1\%$ of the molecules in the solution are glucose.

The Master Equation

Enter Raoult's Law. It elegantly states that the relative lowering of vapour pressure is exactly equal to the mole fraction of the non-volatile solute. Mathematically, it looks like this:

Here, $p^{\circ }$ is the vapour pressure of pure water ($760\text{ torr}$), and $p_s$ is the vapour pressure of our new solution. Let's rearrange this to find the exact drop in pressure ($\Delta p$):

The glucose molecules successfully blocked enough water to drop the pressure by $7.6\text{ torr}$. To find the final vapour pressure of the solution, we simply subtract this drop from the original pure pressure:

And there we have it! By understanding the physical reality of molecules competing for surface space, the mathematics of Raoult's Law becomes a beautiful, logical story rather than just a formula to memorize.