Sulphuric acid reacts with sodium hydroxide as follows: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ When 1 L of 0.1M sulphuric acid solution is allowed to react with 1 L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is (Note: this is a multiple-correct question; two or more options may be correct.) (i) $0.1\ \text{mol L}^{-1}$ (ii) 7.10 g (iii) $0.025\ \text{mol L}^{-1}$ (iv) 3.55 g

Sulphuric acid reacts with sodium hydroxide as follows: H₂SO₄ + 2NaOH arrow Na₂SO₄ + 2H₂O When 1 L of 0.1M sul

NCERT Exemplar · Q17

H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O

(i)

0.1\ \text{mol L}^{-1}

(ii) 7.10 g

(iii)

0.025\ \text{mol L}^{-1}

(iv) 3.55 g

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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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$NaOH$ is the limiting reagent ( $0.1$ mol available vs. $0.2$ mol required), so $0.05$ mol of $Na_2SO_4$ forms $= 7.10$ g, at a molarity of $0.05\ \text{mol}/2\ \text{L}=0.025\ \text{mol L}^{-1}$ . The correct options are (ii) and (iii).

Moles of each reactant

n_{H_2SO_4}=0.1\ \text{M}\times1\ \text{L}=0.1\ \text{mol},\qquad n_{NaOH}=0.1\ \text{M}\times1\ \text{L}=0.1\ \text{mol}

Limiting reagent

The equation needs $2$ mol $NaOH$ per mol $H_2SO_4$ , so $0.1$ mol $H_2SO_4$ would require $0.2$ mol $NaOH$ . Only $0.1$ mol $NaOH$ is present, so $NaOH$ is the limiting reagent.

Moles and mass of $Na_2SO_4$

From the stoichiometry, $2$ mol $NaOH$ give $1$ mol $Na_2SO_4$ :

n_{Na_2SO_4}=\frac{0.1}{2}=0.05\ \text{mol}

With molar mass $M_{Na_2SO_4}=2(23)+32+4(16)=142\ \text{g mol}^{-1}$ :

\text{mass}=0.05\times142=7.10\ \text{g}\quad\Rightarrow\quad\textbf{(ii)}

Molarity in the final solution …

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Let’s break this down step-by-step — first the concept, then the common mistakes, and finally how to avoid each.

The Core Concept: Limiting Reagent & Molarity After Reaction

The balanced equation:

H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O

1 mole of $H_2SO_4$ reacts with 2 moles of $NaOH$ .
We have equal volumes (1 L each) and equal molarities (0.1 M each).
Moles of $H_2SO_4$ = $0.1 \times 1 = 0.1$ mol
Moles of $NaOH$ = $0.1 \times 1 = 0.1$ mol

But the stoichiometry requires twice as much NaOH as $H_2SO_4$ .

So NaOH is the limiting reagent.

Step-by-Step Correct Calculation

From 0.1 mol NaOH, moles of $Na_2SO_4$ formed = $\frac{0.1}{2} = 0.05$ mol
Molar mass of $Na_2SO_4$ = $2(23) + 32 + 4(16) = 142$ g/mol
Mass of $Na_2SO_4$ = $0.05 \times 142 = 7.10$ g
Total volume after mixing = $1 + 1 = 2$ L
Molarity of $Na_2SO_4$ = $\frac{0.05}{2} = 0.025$ mol/L

Correct options: (ii) 7.10 g and (iii) $0.025\ \text{mol L}^{-1}$

Common Mistakes & How to Avoid Each

✗ Mistake 1: Assuming both reactants are completely used (no limiting reagent check)

What students do: They see equal volumes and equal molarities and think both react fully. They then calculate $Na_2SO_4$ from $H_2SO_4$ directly: $0.1$ mol $H_2SO_4$ → $0.1$ mol $Na_2SO_4$ → mass = $14.2$ g, molarity = $0.05$ M.
Why it’s wrong: The equation shows 1:2 ratio, not 1:1. NaOH runs out first.
How to avoid: Always compare mole ratios — not just volumes or molarities. Write the balanced equation and check which reactant gives the smaller product amount.

✗ Mistake 2: Forgetting to add volumes for final molarity

What students do: They calculate molarity as $\frac{0.05}{1} = 0.05$ M (using only one solution’s volume).
Why it’s wrong: After mixing, the total volume is 2 L, not 1 L.
How to avoid: Whenever two aqueous solutions are mixed, total volume = sum of individual volumes (unless stated otherwise). Always divide moles of product by final total volume.

✗ Mistake 3: Using wrong molar mass for $Na_2SO_4$

What students do: They forget to multiply atomic masses correctly — e.g., using 23 for Na but forgetting there are 2 atoms.
Why it’s wrong: Leads to mass like 3.55 g (half of correct) or other wrong numbers.
How to avoid: Write the formula clearly: $Na_2SO_4$ → 2 Na, 1 S, 4 O. Calculate stepwise: …

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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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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? …

Concept: Limiting Reagent & Molarity in Mixture

Steps

The Core Concept: Limiting Reagent & Molarity After Reaction

Step-by-Step Correct Calculation

Common Mistakes & How to Avoid Each

✗ Mistake 1: Assuming both reactants are completely used (no limiting reagent check)

✗ Mistake 2: Forgetting to add volumes for final molarity

✗ Mistake 3: Using wrong molar mass for $Na_2SO_4$

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