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

Exercises · 1.17

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.

Telangana TsbieTextbookSubjective· 2mImportance★★★★★

32% · 42/131 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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? …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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

TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$0.892\,N$
›Reveal solutionSolution

" $20$ volume" means $1\ \text{L}$ of solution releases $20\ \text{L}$ of O $_2$ at STP; using $N=\dfrac{\text{volume strength}}{5.6}$ gives $N=\dfrac{20}{5.6}=3.57\ N$ - option (D).

Meaning of volume strength. A " $20$ volume" H $_2$ O $_2$ solution liberates $20\ \text{L}$ of O $_2$ (at STP) per litre of solution on decomposition:
$2\,\text{H}_2\text{O}_2\rightarrow2\,\text{H}_2\text{O}+\text{O}_2$
Step 1 - Moles of O $_2$ per litre.
$n_{\text{O}_2}=\frac{20}{22.4}=0.893\ \text{mol}$
Step 2 - Moles and equivalents of H $_2$ O $_2$ . …
Start your 14-day free trial to see the full solution →
TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$_3$
›Reveal solutionSolution
The key is to use the balanced chemical equation and stoichiometry to find the mass of pure HCl needed, then convert to the mass of the 50% w/w solution. The required mass is 292 g, so option (B) is correct.

Concept & Intuition

This problem is about reacting hydrochloric acid with calcium carbonate. The reaction is a classic acid–carbonate neutralization:
$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
We are given a 50% w/w solution — meaning 50 g of pure HCl per 100 g of solution. So the actual mass of solution needed will be double the mass of pure HCl required. The trap is forgetting to account for the dilution and just picking the mass of pure HCl.

Step-by-step solution

Write the balanced equation

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
This tells us: 1 mole of CaCO₃ reacts with 2 moles of HCl.

Find moles of CaCO₃ Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol. Given mass = 200 g.

$\text{Moles of CaCO}_3 = \frac{200}{100} = 2 \text{ mol}$

Find moles of HCl needed

From the equation: 1 mol CaCO₃ needs 2 mol HCl.

So 2 mol CaCO₃ need $2 \times 2 = 4$ mol HCl.

Find mass of pure HCl required

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.

$\text{Mass of pure HCl} = 4 \times 36.5 = 146 \text{ g}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
The "volume strength" of hydrogen peroxide is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution the strength (w/v) is about 15%, so the correct option is (D).

Concept & Intuition

"Volume strength" (e.g. "10 volume," "50 volume") labels a hydrogen peroxide solution by the volume of oxygen it releases: 1 mL of the solution produces that many mL of $\mathrm{O_2}$ at STP on decomposition. The decomposition is
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
To convert volume strength into a percentage (g of $\mathrm{H_2O_2}$ per 100 mL of solution), use the molar volume at STP (22.4 L/mol) and the molar mass of $\mathrm{H_2O_2}$ (34 g/mol).

Step-by-step reasoning

Interpret "50 volume"

1 mL of solution yields 50 mL of $\mathrm{O_2}$ at STP, so 1 L of solution yields $50 \times 1000 = 50\,000$ mL = 50 L of $\mathrm{O_2}$ .

Moles of $\mathrm{O_2}$

$n_{\mathrm{O_2}} = \frac{50}{22.4} \approx 2.232\ \text{mol}$

Moles of $\mathrm{H_2O_2}$

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times 2.232 = 4.464$ mol.

Mass of $\mathrm{H_2O_2}$ per litre

With molar mass 34 g/mol,

$m = 4.464 \times 34 \approx 151.8\ \text{g per litre}$

Express as a percentage (w/v) …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
"Volume strength" of $\mathrm{H_2O_2}$ is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution, converting through the decomposition $2\mathrm{H_2O_2} \to 2\mathrm{H_2O} + \mathrm{O_2}$ gives a strength of about 15%, so the correct option is (D).

Why this approach works

"Volume strength" tells you how many millilitres of oxygen gas (at STP) one millilitre of the solution releases on decomposition. So a "50 volume" solution means 1 mL of solution yields 50 mL of $\mathrm{O_2}$ .

To convert this into a percentage by mass (w/v), we:

find the mass of $\mathrm{H_2O_2}$ that produces that volume of oxygen, using the decomposition stoichiometry;

relate that mass to the mass of solution (density approx 1 g/mL for dilute solutions).

The bridge is the balanced equation:
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
so 2 moles of $\mathrm{H_2O_2}$ give 1 mole of $\mathrm{O_2}$ .

Step-by-step reasoning

Moles of oxygen from the given volume At STP, 1 mole of gas occupies 22400 mL. For 50 mL of $\mathrm{O_2}$ (from 1 mL of solution):

$n_{\mathrm{O_2}} = \frac{50}{22400} = \frac{1}{448}\ \text{mol}$

Moles of $\mathrm{H_2O_2}$ that produce this oxygen

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times \frac{1}{448} = \frac{1}{224}\ \text{mol}$ .

Mass of $\mathrm{H_2O_2}$

Molar mass of $\mathrm{H_2O_2} = 34$ g/mol, so

$m = \frac{1}{224}\times 34 = \frac{34}{224} \approx 0.152\ \text{g}$ …
Start your 14-day free trial to see the full solution →

←Previous question1.16 — Heptane and octane form an ideal solution. At 373 K, the vapour pressu…→Next question1.18 — Calculate the mass of a non-volatile solute (molar mass 40 g mol⁻¹) wh…

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

Why this approach works

Step-by-step reasoning

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

🎓Unlock everything free for 14 days

Unlock everything free for 14 days