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

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

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Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

Imagine you're making a cup of sweet tea. You add sugar to hot water, stir, and taste. If you let the tea cool to room temperature, the amount of sugar hasn't changed — but the volume of the liquid has shrunk slightly. If you measured concentration as "grams of sugar per litre of solution," that number would change just because the temperature changed. That's annoying if you're a chemist who needs a reliable, temperature-independent way to describe how much solute is present.

Molality was invented to solve exactly this problem.

The Intuition

Instead of measuring the volume of the solution (which expands and contracts with temperature), molality measures the mass of the solvent. Mass doesn't change with temperature. So molality gives you a concentration that stays the same whether your solution is hot or cold.

Think of it this way:

Molarity = moles of solute per litre of solution (temperature-sensitive)
Molality = moles of solute per kilogram of solvent (temperature-independent)

The solvent is the substance doing the dissolving — usually water. The solute is what gets dissolved — sugar, salt, etc.

The Precise Definition

\text{Molality } (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

The symbol for molality is a lowercase m (not to be confused with M for molarity).

Key points to remember:

The denominator is solvent mass, not solution mass
The unit is mol/kg (often written as simply "m")
It is independent of temperature because mass doesn't change with temperature

Worked Example

Problem: 36 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.

Step 1: Find moles of solute

\text{Moles of glucose} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}

Step 2: Convert solvent mass to kilograms

500 \text{ g} = 0.5 \text{ kg}

Step 3: Apply the formula

m = \frac{0.2 \text{ mol}}{0.5 \text{ kg}} = 0.4 \text{ m}

Note

The answer is 0.4 m (or 0.4 mol/kg). Notice we used the mass of water (500 g), not the mass of the solution (which would be 536 g).

Common Mistake to Avoid

Watch out

Do not use the mass of the solution in the denominator. The formula specifically asks for the mass of the solvent alone. If the problem gives you the total mass of the solution, subtract the mass of the solute to find the solvent mass.

When Do You Use Molality?

Molality is the star in two important situations: …

Why this formula?

Molality Calculation: Why the Formula Works

Molality is a measure of concentration that is temperature-independent — this is its key advantage over molarity. Let's understand why the formula takes the form it does.

The Definition First

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

The unit is mol/kg, often written as m (e.g., 0.5 m glucose solution).

Why Mass of Solvent, Not Solution?

This is the critical conceptual point.

The Reasoning

Molarity uses volume of solution → volume changes with temperature (expansion/contraction). So molarity changes with temperature.
Molality uses mass of solvent → mass is invariant with temperature. So molality remains constant regardless of temperature changes.

Key insight: By using the solvent's mass (not the solution's volume), we eliminate temperature dependence. This is why molality is preferred for colligative properties (boiling point elevation, freezing point depression) — these properties depend on the number of solute particles, not on temperature.

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

If you have $w_{\text{solute}}$ grams of solute with molar mass $M_{\text{solute}}$ (g/mol):

\text{moles of solute} = \frac{w_{\text{solute}}}{M_{\text{solute}}}

Step 2: Mass of Solvent in kg

If the solvent mass is $W_{\text{solvent}}$ grams:

\text{mass of solvent in kg} = \frac{W_{\text{solvent}}}{1000}

Step 3: Putting It Together

m = \frac{\frac{w_{\text{solute}}}{M_{\text{solute}}}}{\frac{W_{\text{solvent}}}{1000}}

Simplifying:

m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}

The Final Formula (Exam-Ready)

\boxed{m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}}

Where:

$w_{\text{solute}}$ = mass of solute in grams
$M_{\text{solute}}$ = molar mass of solute in g/mol
$W_{\text{solvent}}$ = mass of solvent in grams

Why the ×1000 Factor? …

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The vapour pressure of an aqueous glucose solution is lower than that of pure water because the non-volatile glucose molecules occupy space at the liquid surface, reducing the number of water molecules that can escape into the vapour phase — this is Raoult’s law for a non-volatile solute.

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.

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.
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.
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 ( $x_{\text{solvent}}$ ):

P = x_{\text{solvent}} \cdot P^0

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

P_{\text{solution}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^0

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.
A common misconception to avoid: …

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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." …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$\mathrm{CH_3OH}$
›Reveal solutionSolution
Convert mole fraction of solute directly to molality using $m=\dfrac{x_2}{x_1}\cdot\dfrac{1000}{M_1}$ , taking $M_1=18\ \mathrm{g/mol}$ for water. Answer: 1.13 m.

Concept and Intuition

Molality ( $m$ = moles solute per kg solvent) and mole fraction ( $x_2$ = moles solute per total moles) are both intensive composition measures, so one converts to the other purely through mole-count and molar-mass arithmetic — no need to assume any solution volume or density, which is exactly why molality/mole-fraction problems are solvable without extra data (unlike molarity, which needs density).

Step-by-Step Solution

Given $x_2$ (mole fraction of $\mathrm{CH_3OH}$ ) $=0.02$ , so $x_1$ (mole fraction of water) $=1-0.02=0.98$ .

Take a basis of 1 mole total solution: moles of $\mathrm{CH_3OH}$ , $n_2=0.02$ ; moles of water, $n_1=0.98$ .

Mass of water (solvent) $= n_1 \times M_1 = 0.98 \times 18\ \mathrm{g/mol} = 17.64\ \mathrm{g} = 0.01764\ \mathrm{kg}$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$g\ mL^{-1}$
›Reveal solutionSolution
Find the NaOH needed to neutralize 0.5 mol HCl (= 0.5 mol NaOH), back out the water present at $x_{NaOH}=0.02$ , and convert that water's mass to volume using its density — giving ≈441 mL.

Concept and Intuition

Mole fraction directly links the moles of solute to the moles of solvent present. Once we know exactly how many moles of NaOH must be present (fixed by the stoichiometric neutralization requirement), the mole-fraction relation pins down exactly how much water accompanies it — and hence, via water's known density, the volume of solution.

Step-by-Step Solution

Moles of HCl = $1\ L \times 0.5\ mol/L = 0.5$ mol.

$NaOH + HCl \rightarrow NaCl + H_2O$ is 1:1, so moles of NaOH required = 0.5 mol.

Mole fraction: $x_{NaOH} = \dfrac{n_{NaOH}}{n_{NaOH}+n_{H_2O}} = 0.02$ .

$\Rightarrow n_{NaOH}+n_{H_2O} = \dfrac{0.5}{0.02} = 25 \Rightarrow n_{H_2O} = 24.5$ mol.

Mass of water = $24.5 \times 18\ g/mol \approx 441$ g. …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$Q.A solid solute is dissolved in water. The mole fraction of solute is 0.02. What is the molality of the solution? (A) 2.133 m (B) 2.5 m (C) 1.5 m (D) 1.133 m$
›Reveal solutionSolution
Convert mole fraction to molality by picking a convenient basis (1 mol total) and computing the solvent's mass. Answer: 1.133 m.

Concept and Intuition

Mole fraction and molality are related but different concentration scales: mole fraction is a ratio of moles, while molality is moles of solute per kilogram of solvent. To convert, assume a convenient total amount (1 mole of solution), find the moles of each component from the given mole fraction, convert the solvent's moles to a mass, and then compute molality directly.

Step-by-Step Solution

Given: mole fraction of solute $x_{solute}=0.02$ , so mole fraction of water $x_{water}=1-0.02=0.98$ .

Assume 1 mol of total solution: $n_{solute}=0.02$ mol, $n_{water}=0.98$ mol.

Mass of water $= n_{water}\times M_{water} = 0.98\times18 = 17.64$ g $= 0.01764$ kg. …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$\mathrm{C_2H_6O_2}$
›Reveal solutionSolution
This is a direct molality calculation: moles of solute per kilogram of solvent.

Concept and Intuition

Molality is defined as moles of solute dissolved per kilogram of solvent (not solution), making it temperature-independent and ideal for colligative-property calculations like antifreeze formulations. Ethylene glycol's molar mass must first be computed from its formula to find the moles present.

Step-by-Step Solution

Molar mass of $C_2H_6O_2$ : $2(12)+6(1)+2(16) = 24+6+32 = 62\ \mathrm{g\,mol^{-1}}$ .

Moles of ethylene glycol $= \dfrac{248}{62} = 4$ mol.

Mass of water (solvent) $= 200\ \mathrm{g} = 0.200\ \mathrm{kg}$ . …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
Molality uses moles of solute per kg of solvent (not solution). For 10% w/w glucose, 100 g solution = 10 g glucose + 90 g water, giving molality $\approx 0.62$ m.

Concept and Intuition

Molality is defined relative to the mass of solvent only, unlike mass percent (w/w) which is defined relative to total solution mass. So the first step is always to extract the solvent mass from the given composition before applying $m = \dfrac{n_{solute}}{\text{mass of solvent (kg)}}$ .

Step-by-Step Solution

Basis: 100 g of solution. 10% (w/w) glucose means 10 g glucose and $100-10=90$ g water.

Moles of glucose $= \dfrac{10\text{ g}}{180\text{ g mol}^{-1}} = 0.05\overline{5}$ mol.

Mass of solvent (water) in kg $= 90\text{ g} = 0.090$ kg. …
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AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$= 62\ \mathrm{g\ mol^{-1}}$
›Reveal solutionSolution
This is a direct molality calculation. 124 g of ethylene glycol is 2 mol; requiring a 10 m solution fixes the water mass at 200 g.

Concept and Intuition

Molality (m) is defined per kilogram of solvent, not solution — it is temperature-independent and depends only on moles of solute and mass of solvent.
$m = \frac{n_{solute}}{w_{solvent}(\text{kg})}$
Step-by-Step Solution

Moles of ethylene glycol $= \dfrac{124\ \text{g}}{62\ \text{g mol}^{-1}} = 2\ \text{mol}$ .

Let mass of water $= x$ g $= \dfrac{x}{1000}$ kg.

Given molality $= 10$ m: $10 = \frac{2}{x/1000}$ …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\mathrm{dm^3}$
›Reveal solutionSolution
Converting molarity to molality requires the mass of solvent (not solution), obtained by subtracting the solute's mass from the total solution mass computed via density.

Concept and Intuition

Molarity (mol/L of solution) and molality (mol/kg of solvent) are only related once you know the density of the solution, because you need to convert the solution's volume to solution mass and then subtract the solute mass to get solvent mass.

Step-by-Step Solution

Molarity: $x = \dfrac{82\ \mathrm{g}}{60\ \mathrm{g/mol}} = 1.3\overline{6}$ mol per litre of solution (so moles of acetic acid per litre solution $= x$ ).

Mass of 1 L (1 dm³) of solution $= 1.01\ \mathrm{kg/dm^3} \times 1\ \mathrm{dm^3} = 1.01\ \mathrm{kg} = 1010$ g.

Mass of solvent (water) $= 1010 - 82 = 928$ g $= 0.928$ kg. …
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AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
$\mathrm{Na_2CO_3}$
›Reveal solutionSolution
Direct molality calculation: moles of solute divided by kilograms of solvent gives 0.02 m.

Concept and Intuition

Molality is defined as moles of solute per kilogram of solvent (not solution), making it temperature-independent and convenient for colligative property calculations.

Step-by-Step Solution

Moles of $\mathrm{Na_2CO_3} = \dfrac{1.06\ \mathrm{g}}{106\ \mathrm{g\,mol^{-1}}} = 0.01$ mol.

Mass of solvent (water) = 500 g = 0.500 kg.

Molality $= \dfrac{0.01\ \mathrm{mol}}{0.500\ \mathrm{kg}} = 0.02$ mol/kg = 0.02 m. …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$Q. The molarity of one molal glucose solution having density of 1.2 g/mL is (A) 0.101 M (B) 1.01 M (C) 2.01 M (D) 0.001 M$
›Reveal solutionSolution
Convert 1 molal glucose solution to molarity using the solution's density; molality and molarity differ because molarity is based on total solution volume, not solvent mass. Answer: 1.01 M.

Concept and Intuition

Molality (mol/kg solvent) and molarity (mol/L solution) are numerically close for dilute aqueous solutions but not identical, because molality ignores the solute's contribution to the total solution mass/volume. Density lets us convert between the two using $\text{Molarity} = \dfrac{1000 \times m \times d}{1000 + m\times M_w}$ , where $m$ is molality, $d$ is density (g/mL), and $M_w$ is the solute's molar mass.

Step-by-Step Solution

Take 1 kg (1000 g) of water as solvent, containing 1 mol glucose ( $M_w = 180\ \mathrm{g/mol}$ , so mass = 180 g).

Total solution mass $= 1000 + 180 = 1180$ g. …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$\mathrm{H_2O}$
›Reveal solutionSolution
Molality is moles of solute per kilogram of solvent; with 0.1 mol glucose in 0.018 kg water, molality $=5.55\,\text{m}$ .

Concept and Intuition

Molality is defined using the mass of solvent (not volume, and not total solution mass), which is exactly why it is temperature-independent and preferred in colligative-property calculations. Here both the solute (glucose) and the solvent (water) are given in grams, so both must first be converted appropriately — glucose to moles, water to kilograms.

Step-by-Step Solution

Moles of glucose ( $M=180\,\text{g/mol}$ ): $n = \dfrac{18\,\text{g}}{180\,\text{g/mol}} = 0.1\,\text{mol}$ .

Mass of water in kg: $18\,\text{g} = 0.018\,\text{kg}$ . …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
Converting 15 ppm (by mass) of a solute of molar mass 120 g/mol into molality gives $1.25\times10^{-4}$ mol kg $^{-1}$ .

Concept and Intuition

ppm (parts per million) by mass means grams of solute per million grams of solution; for a dilute aqueous solution, the solution mass is essentially the water mass. Molality is moles of solute per kilogram of solvent, so we convert the mass basis to moles and then to per-kg terms.

Step-by-Step Solution

15 ppm by mass $\Rightarrow$ 15 g of carcinogen per $10^6$ g ( $=1000$ kg) of water.

Moles of carcinogen $= \dfrac{15\text{ g}}{120\text{ g/mol}} = 0.125$ mol. …
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AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$\mathrm{mL^{-1}}$
›Reveal solutionSolution
Converts a 10% w/w NaOH solution into molarity using its density and molar mass.

Concept and Intuition

Working with exactly 1 L (1000 mL) of solution makes the arithmetic simplest: the total mass of that litre comes from its density, 10% of that mass is NaOH, and dividing by NaOH's molar mass gives the number of moles present in that one litre — which is the molarity.

Step-by-Step Solution

Mass of 1 L (1000 mL) of solution $=1000\times1.11=1110$ g.

Mass of NaOH in it (10% w/w) $=0.10\times1110=111$ g.

Moles of NaOH $=\dfrac{111}{40}=2.775$ mol. …
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Q.Why is the vapour pressure of an aqueous solution of glucose lower than that of water?

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