Animated Solution for Chemistry - Solutions: 18 g of glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at $100^{\circ }\text{C}$ is

## The Invisible Shield: How Glucose Lowers Vapour Pressure

Imagine you are standing in a kitchen, watching a beaker of pure water boiling vigorously. The steam is rising, and the water molecules are escaping into the air with immense energy. This simple observation holds the key to solving our problem.

The question states that the temperature of the aqueous solution is $100^{\circ }\text{C}$. This is a massive, hidden hint! We know that $100^{\circ }\text{C}$ is the normal boiling point of pure water. At its boiling point, the vapour pressure of any liquid becomes exactly equal to the external atmospheric pressure.

Therefore, the vapour pressure of pure water, denoted as $p^{\circ }$, is $1\text{ atm}$, which is equivalent to $760\text{ torr}$.

Now, we introduce a twist: we add $18\text{ g}$ of glucose (${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$) to the water. Glucose is a non-volatile solute. It doesn't vaporize, but its bulky molecules occupy space at the surface of the liquid, acting like an invisible shield that blocks water molecules from escaping. Consequently, the vapour pressure of the solution drops.

To quantify this drop, we invoke Raoult's Law for the relative lowering of vapour pressure:

Here, $p_s$ is the vapour pressure of the solution, and $x_{\text{solute}}$ is the mole fraction of the glucose.

Before we can use Raoult's Law, we need to find the mole fraction of glucose. Let's calculate the moles of both components.

First, the solute (glucose): - Mass of glucose $=18\text{ g}$ - Molar mass of glucose $=180\text{ g/mol}$ - Moles of glucose ($n$) $=\frac{18}{180}=0.1\text{ mol}$

Next, the solvent (water): - Mass of water $=178.2\text{ g}$ - Molar mass of water $=18\text{ g/mol}$ - Moles of water ($N$) $=\frac{178.2}{18}=9.9\text{ mol}$

The total number of moles in the solution is $n+N=0.1+9.9=10\text{ mol}$.

Now, the mole fraction of glucose ($x_{\text{solute}}$) is simply its share of the total moles:

We have all the pieces of the puzzle. Let's substitute them back into Raoult's Law:

Multiplying both sides by $760$ gives:

Rearranging to solve for $p_s$:

And there we have it! By adding just a small amount of glucose, we successfully lowered the vapour pressure of the water from $760\text{ torr}$ to $752.40\text{ torr}$. This elegant calculation perfectly demonstrates the predictable nature of colligative properties.