The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-vola

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Q. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

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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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A non-volatile solute lowers the vapour pressure of water through Raoult's law; for a 1 molal aqueous solution, the mole fraction of water is approximately 0.982, giving a vapour pressure of 12.08 kPa.

When a non-volatile solute dissolves in a solvent, it occupies space at the surface and reduces the number of solvent molecules that can escape into the vapour phase. This phenomenon—vapour pressure lowering—is quantified by Raoult's law, which states that the vapour pressure of the solution is proportional to the mole fraction of the solvent.

The key insight is that molality tells us moles of solute per kilogram of solvent, so we can convert this to mole fractions and then apply Raoult's law directly.

P_{\text{solution}} = \chi_{\text{solvent}} \cdot P^\circ_{\text{solvent}}

where $\chi_{\text{solvent}}$ is the mole fraction of the solvent and $P^\circ_{\text{solvent}}$ is the vapour pressure of the pure solvent.

Step-by-step calculation

Identify what 1 molal means.

A 1 molal solution contains 1 mole of solute dissolved in 1 kg (1000 g) of water. We need to find how many moles of water that corresponds to.
Calculate moles of water.

The molar mass of water is 18 g/mol, so:

n_{\text{water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol}

Find the mole fraction of water. The total moles in solution = moles of water + moles of solute = $55.56 + 1 = 56.56$ mol.

\chi_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{solute}}} = \frac{55.56}{56.56} = 0.9823

Apply Raoult's law. …

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Vapour Pressure of a 1 Molal Solution – Raoult’s Law Application

1. Concept First – The Idea Being Tested

This question tests Raoult’s Law for a non-volatile solute. The core idea is simple:

When you dissolve a non-volatile solute (one that doesn’t evaporate) in a solvent, the solute particles block some solvent molecules from escaping into the vapour phase. This lowers the vapour pressure of the solvent.

Why does this happen?

In pure water, all surface molecules can evaporate freely.
When solute is added, some surface sites are occupied by solute particles (which don’t evaporate).
Fewer water molecules are available at the surface to escape → vapour pressure drops.

Raoult’s Law quantifies this:

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

Where:

$P_{\text{solution}}$ = vapour pressure of the solution
$X_{\text{solvent}}$ = mole fraction of the solvent
$P_{\text{solvent}}^0$ = vapour pressure of pure solvent (given as 12.3 kPa)

The key challenge here: we are given molality (1 molal), not mole fraction. So we must convert.

2. Step-by-Step Solution

Step 1: Understand what “1 molal” means

A 1 molal solution means:

1 mole of solute dissolved in 1 kg of solvent (water).

So we have:

Moles of solute, $n_{\text{solute}} = 1 \text{ mol}$
Mass of water = 1000 g

Step 2: Find moles of water (solvent)

Molar mass of water ( $H_2O$ ) = 18 g/mol

n_{\text{water}} = \frac{\text{mass of water}}{\text{molar mass}} = \frac{1000}{18} = 55.56 \text{ mol}

Step 3: Calculate mole fraction of water

Mole fraction of solvent (water) is:

X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{solute}}}

Substitute:

X_{\text{water}} = \frac{55.56}{55.56 + 1} = \frac{55.56}{56.56}

Calculate:

X_{\text{water}} = 0.9823

Why this step?

Raoult’s Law uses mole fraction, not molality. We must convert because the “blocking effect” depends on the relative number of solute vs solvent particles.

Step 4: Apply Raoult’s Law

P_{\text{solution}} = X_{\text{water}} \cdot P_{\text{water}}^0

Given $P_{\text{water}}^0 = 12.3 \text{ kPa}$ :

P_{\text{solution}} = 0.9823 \times 12.3

P_{\text{solution}} = 12.08 \text{ kPa}

3. Final Answer

The vapour pressure of the 1 molal solution is $\boxed{12.08 \text{ kPa}}$ .

4. Why It Works & Exam Tip

Why it works …

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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 31 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$2.0$
›Reveal solutionSolution
A 2.0 molal solution means 2 moles of solute in 1 kg of water. Convert the solvent mass to moles, then apply the mole fraction formula: $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$ . The answer is $\boxed{0.0347}$ .

Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. This distinction matters because we need to count the moles of both solute and solvent separately to find the mole fraction.

When we say a solution is 2.0 molal, we're saying there are 2.0 moles of solute dissolved in exactly 1000 g (1 kg) of water. The mole fraction then asks: what fraction of the total number of particles (molecules) in the solution comes from the solute?

Let me work through the calculation systematically.

1. Identify what we know from "2.0 molal aqueous solution"

The molality $m = 2.0$ tells us:

Moles of solute: $n_{\text{solute}} = 2.0 \text{ mol}$

Mass of water (solvent): $1000 \text{ g}$

2. Convert the mass of water to moles

Water has a molar mass of $18 \text{ g/mol}$ , so:
$n_{\text{water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol}$
3. Calculate the total moles in the solution
$n_{\text{total}} = n_{\text{solute}} + n_{\text{water}} = 2.0 + 55.56 = 57.56 \text{ mol}$
4. Apply the mole fraction formula

The mole fraction of solute is:

$\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{total}}} = \frac{2.0}{57.56} = 0.03474$ …
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CBSE 2026Set A1 markMCQ
$Q. 34.2 g of sugar is present in 234.2 g of its aqueous solution. Then its molal concentration is$
(a) 0.1
(b) 0.5
(c) 5.5
(d) 55.0
›Reveal solutionSolution

Moles of sugar = 34.2/342 = 0.1; mass of solvent (water) = 234.2 - 34.2 = 200 g = 0.2 kg; molality = 0.1/0.2 = 0.5 m.

Molar mass of sugar (sucrose) = 342 g/mol.

Moles of sugar = 34.2/342 = 0.1 mol. …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. What will be the molarity of 30 ml of 0.5 M H2SO4 solution diluted to 50 ml?$
(a) 0.3 M
(b) 0.03 M
(c) 3 M
(d) 0.13 M
›Reveal solutionSolution

Dilution does not change the number of moles of solute, so M1V1 (before) = M2V2 (after).

Given: M1 = 0.5 M, V1 = 30 mL, final volume V2 = 50 mL.

…

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CBSE 2026Set ANNUAL1 mark
$Q.Define mole fraction.$
›Reveal solutionSolution

Mole fraction expresses a component's amount relative to the total moles in a mixture, independent of temperature.

For a solution containing components with n1, n2, n3, ... moles, the mole fraction of component 1 is defined as:

x1 = n1 / (n1 + n2 + n3 + ...)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In an acid-base titrimetric analysis, the concentration of a sulphuric acid analyte is found to be 0.044 M. The strength of the acid in g/L is -$
(a) 0.44
(b) 4.31
(c) 2.15
(d) 44.00
›Reveal solutionSolution

Strength in g/L = molarity × molar mass = 0.044 × 98 ≈ 4.31 g/L, so option (B).

The strength of a solution in grams per litre is related to its molarity by

$\text{Strength (g/L)} = \text{Molarity (mol/L)} \times \text{Molar mass (g/mol)}.$

…

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CBSE 2025Set ANNUAL1 mark
$Q. Write the definition of molality.$
›Reveal solutionSolution

Molality expresses concentration as moles of solute per kilogram of SOLVENT (not solution), and unlike molarity it does not change with temperature.

Definition:

Molality (m) = (moles of solute) / (mass of solvent in kg)

Unit: mol kg^-1 (also written 'molal', symbol m)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the molarity of a solution with a mass of solute 10 kg mass and 100 litre volume?$
(a) 0.1 molar
(b) 1 molar
(c) 10 molar
(d) 100 molar
›Reveal solutionSolution

Molarity is defined as moles of solute per litre of solution; dividing the given amount of solute by the solution volume gives 0.1 M.

Molarity is defined as:

$M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$

Note: as printed, the question states the solute quantity as '10 kg mass'; for the arithmetic to match any of the given options (0.1, 1, 10, 100 M) the intended quantity is 10 moles of solute (a common wording/printing slip in this recurring question, where 'kg' should read 'mol') - with n …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A solution contains 8 moles of solute and the mass of solvent is 4 kg. What is the molality of this solution?$
(a) 5 mol/kg
(b) 8 mol/kg
(c) 4 mol/kg
(d) 2 mol/kg
›Reveal solutionSolution
Molality is moles of solute per kilogram of solvent (not solution); here it works out to 8/4 = 2 mol/kg.

Molality (m) is defined as:

m = (moles of solute) / (mass of solvent in kg)

Given:

moles of solute = 8 mol

mass of solvent = 4 kg

m = 8 mol / 4 kg = 2 mol/kg

…
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CBSE 2025Set ANNUAL1 mark
$Q. Define molality.$
›Reveal solutionSolution

Molality is defined as moles of solute per kilogram of solvent; unlike molarity, it does not depend on temperature since it is based on mass, not volume.

Molality (denoted m) of a solution is defined as the number of moles of solute dissolved in one kilogram (1000 g) of solvent:

Molality (m) = (number of moles of solute) / (mass of solvent in kg)

Unit: mol kg^-1 (also written as 'm', e.g., a '1 molal' or '1 m' solution).

…

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CBSE 2024Set A11 mark
$Q.The number of moles of solute present in one kilogram of the solvent is called \rule{2cm}{0.4pt}.$
›Reveal solutionSolution

Moles of solute per kilogram of solvent defines molality.

Molality ( $m$ ) is a concentration term that depends only on the mass of solvent (and is therefore temperature-independent):

$m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} \;\; (\text{mol kg}^{-1})$ …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following has no unit?$
(a) Molarity
(b) Molality
(c) Normality
(d) Molar Fraction
›Reveal solutionSolution

Mole fraction is a dimensionless ratio, so it has no unit.

…

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CBSE 2024Set ANNUAL1 mark
$Q. Write down the formula of Molarity.$
›Reveal solutionSolution

Molarity is the number of moles of solute dissolved per litre of solution.

Molarity is one of the most widely used units of concentration. It is defined as the number of moles of solute dissolved in one litre (one cubic decimetre) of solution:

$M = \frac{\text{moles of solute } (n)}{\text{volume of solution in litres } (V)}$ …

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Q.The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

The Intuition

The Precise Definition

Worked Example

Common Mistake to Avoid

When Do You Use Molality?

Why this formula?

Molality Calculation: Why the Formula Works

The Definition First

Why Mass of Solvent, Not Solution?

The Reasoning

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

Step 2: Mass of Solvent in kg

Step 3: Putting It Together

The Final Formula (Exam-Ready)

Why the ×1000 Factor? …

Step-by-step calculation

Vapour Pressure of a 1 Molal Solution – Raoult’s Law Application

1. Concept First – The Idea Being Tested

2. Step-by-Step Solution

Step 1: Understand what “1 molal” means

Step 2: Find moles of water (solvent)

Step 3: Calculate mole fraction of water

Step 4: Apply Raoult’s Law

3. Final Answer

4. Why It Works & Exam Tip

Why it works …

1. Using Molality Directly in Raoult's Law

2. Forgetting What "1 Molal" Means

3. Using the Mole Fraction of Solute Instead of Solvent

4. Forgetting That Water's Molar Mass Gives About 55.5 mol per kg

Q.The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Step-by-step calculation

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