Q.Why is the vapour pressure of an aqueous solution of glucose lower than that of water?

The key idea is Raoult's Law: the vapour pressure of a solvent above a solution is proportional to its mole fraction in the solution.

For pure water, the mole fraction of water is 1, so the vapour pressure is $P^0_{\text{water}}$.

When glucose (a non-volatile solute) is dissolved, the mole fraction of water becomes less than 1. According to Raoult's Law:

$P_{\text{solution}} = X_{\text{water}} \cdot P^0_{\text{water}}$ …

The key idea is simple: vapour pressure depends on how easily solvent molecules can leave the liquid surface. In pure water, every molecule at the surface is a water molecule, free to evaporate. When you dissolve glucose — a non-volatile solute (it doesn’t evaporate itself) — some of those surface spots are taken by glucose molecules. They don’t contribute to vapour pressure. So fewer water molecules can escape per unit area, and the vapour pressure drops.

This is not a chemical effect — glucose does not react with water. It’s purely a physical, surface-statistics effect. The more glucose you add, the fewer water molecules at the surface, and the lower the vapour pressure.

Let’s walk through the reasoning step by step.

1. What determines vapour pressure?

   At any temperature, molecules in a liquid have a range of kinetic energies. Some near the surface have enough energy to overcome intermolecular forces and escape into the gas phase. The vapour pressure is the pressure exerted by these escaped molecules when equilibrium is reached. For a pure liquid, this is a fixed value at a given temperature — for water at 25 °C, it’s about 23.8 mm Hg.

2. What changes when glucose is added?

   Glucose ($\ce{C6H12O6}$) dissolves in water but does not evaporate — it’s non-volatile. Once dissolved, glucose molecules are distributed throughout the solution, including at the surface. At the surface, some fraction of the available area is now occupied by glucose molecules. Water molecules can only evaporate from spots where water is at the surface.

3. Raoult’s law gives the quantitative relationship.

   For a solution of a non-volatile solute, the vapour pressure of the solvent above the solution ($P$) is directly proportional to the mole fraction of the solvent ($x_{\text{solvent}}$):

where $P^0$ is the vapour pressure of the pure solvent. Since $x_{\text{solvent}} < 1$ (because some of the moles are glucose), $P < P^0$.

4. Why is this a lowering, not a destruction?

   The vapour pressure doesn’t go to zero (unless you add so much glucose that no water is at the surface — which is impossible because glucose is soluble only up to a limit). It’s simply reduced proportionally to how much of the surface is “blocked” by solute. For a dilute solution, the lowering is small; for a concentrated one, it’s larger.

5. A common misconception to avoid: …

Concept: Raoult's Law and Vapour Pressure Lowering

Raoult's Law states that for a non-volatile solute (like glucose) dissolved in a volatile solvent (like water), the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.

Method: Vapour Pressure Lowering (Raoult's Law)

Step 1: Identify the components

• Solvent: Water (volatile) — its molecules can escape into vapour phase.
• Solute: Glucose (non-volatile) — it does not evaporate.

Step 2: Apply Raoult's Law

For a solution:

Where:

• $P_{\text{solution}}$ = vapour pressure of the solution
• $X_{\text{solvent}}$ = mole fraction of water in the solution
• $P_{\text{solvent}}^0$ = vapour pressure of pure water at the same temperature

Step 3: Compare mole fractions

• In pure water: $X_{\text{water}} = 1$
• In glucose solution: $X_{\text{water}} < 1$ (because glucose molecules occupy some of the space, reducing the fraction of water molecules)

Step 4: Consequence

Since $X_{\text{water}} < 1$, multiplying by $P_{\text{water}}^0$ gives:

Therefore, the vapour pressure of the glucose solution is lower than that of pure water.

Why this happens physically

• At the surface of pure water, all molecules are water — many can escape into vapour. …

Here are the common mistakes students make when explaining why the vapour pressure of an aqueous glucose solution is lower than that of pure water, along with how to avoid each.

Mistake 1: Saying "Glucose evaporates less" or "Glucose has low vapour pressure"

• The Error: Students often think the solute (glucose) is the one contributing less vapour. They might say, "Glucose is non-volatile, so it doesn't evaporate, hence the pressure is lower."
• Why it's wrong: While it's true glucose is non-volatile, the reason the total vapour pressure drops is not simply because glucose doesn't evaporate. The key is that glucose reduces the number of water molecules at the surface.
• How to avoid: Focus on the solvent (water). The vapour pressure of the solution is due only to water molecules escaping. Glucose molecules occupy space at the surface, blocking water molecules from escaping.

Mistake 2: Confusing "Rate of Evaporation" with "Vapour Pressure"

• The Error: Students say, "The rate of evaporation decreases, so vapour pressure decreases." While this is true, it is incomplete. They often stop here without explaining why the rate decreases.
• Why it's wrong: The exam expects the molecular-level mechanism. Simply stating "rate decreases" is a description, not an explanation.
• How to avoid: Always connect the rate decrease to surface blockage. Write: "Glucose molecules occupy a fraction of the liquid surface, reducing the number of water molecules per unit area that can escape into the vapour phase. Hence, the rate of evaporation decreases, leading to a lower equilibrium vapour pressure."

Mistake 3: Forgetting the Dynamic Equilibrium Argument

• The Error: Students only talk about evaporation and ignore condensation. They say, "Less water evaporates, so pressure is lower."
• Why it's wrong: Vapour pressure is an equilibrium property. If only evaporation slows down, the system would not reach a steady state. In reality, the condensation rate also changes.
• How to avoid: Explain both sides:
  • Evaporation: Slows down due to fewer water molecules at the surface.
  • Condensation: Initially, the vapour density is the same as pure water, so condensation rate is higher than the new evaporation rate.
  • Result: More vapour condenses than evaporates until a new, lower equilibrium vapour pressure is reached.

Mistake 4: Using Raoult's Law Without Understanding

• The Error: Students write: "According to Raoult's law, $P = X_{\text{solvent}} P^0$, so since $X_{\text{solvent}} < 1$, pressure is lower." This is mathematically correct but often given without explanation.
• Why it's wrong: The examiner wants to know why Raoult's law holds. Simply quoting the formula is not enough for a "why" question.
• How to avoid: First explain the physical reason (surface blockage), then state Raoult's law as the quantitative summary. For example: "Because glucose molecules reduce the mole fraction of water at the surface, the escaping tendency of water decreases. This is quantitatively given by Raoult's law: $P = X_{\text{water}} P^0_{\text{water}}$."

Mistake 5: Saying "Glucose attracts water molecules" (Wrong Intermolecular Force)

• The Error: Students claim: "Glucose forms hydrogen bonds with water, holding them back, so vapour pressure decreases." …