[Note: two or more options may be correct] Which of the following binary mixtures will have same composition i

NCERT Exemplar · Q32

Q. [Note: two or more options may be correct] Which of the following binary mixtures will have same composition in liquid and vapour phase?

(i) Benzene - Toluene

(ii) Water-Nitric acid

(iii) Water-Ethanol

(iv) n-Hexane - n-Heptane

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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 liquid–vapour mixture has identical composition in both phases only at an azeotropic point, where the system boils without change in composition. Among the options, water–nitric acid and water–ethanol form azeotropes.

When a liquid mixture boils, the vapour is usually richer in the more volatile component—that's the basis of distillation. But certain mixtures reach a special composition where the liquid and vapour have exactly the same makeup. At this point, called an azeotrope, the mixture behaves like a pure substance: it boils at a constant temperature and the vapour has the same mole fractions as the liquid. No amount of simple distillation can separate an azeotropic mixture further.

Azeotropes arise from strong intermolecular interactions between unlike molecules. When these interactions are significantly different from those in the pure components, Raoult's law breaks down and the vapour pressure curve develops an extremum (maximum or minimum). At that extremum, the composition of liquid and vapour coincide.

Now let's examine each pair:

Benzene–Toluene (i)

Both are non-polar aromatic hydrocarbons with very similar structures and intermolecular forces (London dispersion). They form an ideal solution that obeys Raoult's law closely across all compositions. The vapour is always enriched in benzene (the more volatile component), so liquid and vapour compositions differ at every point. No azeotrope forms.
Water–Nitric acid (ii)

This system exhibits strong hydrogen bonding and significant deviations from ideality. Water and nitric acid form a maximum-boiling azeotrope at approximately 68% (by mass) $\ce{HNO3}$ and 120.5 °C. At this composition, liquid and vapour have the same concentration.
Water–Ethanol (iii)

Ethanol and water hydrogen-bond with each other, but the ethanol–water interaction is weaker than water–water hydrogen bonds. This leads to positive deviation from Raoult's law and a minimum-boiling azeotrope at roughly 95.6% ethanol (by mass) and 78.2 °C. At the azeotropic point, both phases share the same composition—this is why you cannot distill absolute (100%) ethanol from a water–ethanol mixture by simple distillation. …

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Concept: Azeotropes

An azeotrope is a mixture of two liquids that boils at a constant temperature and has the same composition in the liquid and vapour phase. This happens when the mixture's boiling point is either a minimum or a maximum on the boiling point-composition diagram.

Method: Identify Azeotrope-Forming Mixtures

Step 1: Recall the types of binary mixtures

Ideal mixtures (Raoult's law obeyed): No azeotrope. Composition in liquid and vapour is always different (except at pure components).
Non-ideal mixtures with large deviations: Can form azeotropes.

Step 2: Check each option

Option	Mixture	Behaviour	Forms azeotrope?
(i)	Benzene - Toluene	Ideal solution (similar structure, obey Raoult's law)	✗ No
(ii)	Water - Nitric acid	Maximum boiling azeotrope (strong H-bonding, negative deviation)	✓ Yes
(iii)	Water - Ethanol	Minimum boiling azeotrope (positive deviation from Raoult's law)	✓ Yes

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Common Mistakes & How to Avoid Them

Mistake 1: Assuming all ideal solutions have same composition in both phases

Why it's wrong:

Only azeotropic mixtures have identical liquid and vapour composition at a specific boiling point. Ideal solutions (like Benzene-Toluene, n-Hexane-n-Heptane) follow Raoult's law perfectly — their vapour is always richer in the more volatile component.

How to avoid:

Remember:

Ideal solutions → vapour composition ≠ liquid composition (except at pure component ends)
Azeotropes → at a specific composition, vapour = liquid composition

Mistake 2: Confusing minimum-boiling azeotropes with maximum-boiling ones

Why it's wrong:

Students often think all non-ideal solutions form azeotropes, but they forget the type:

Water-Ethanol → minimum-boiling azeotrope (vapour richer in ethanol below azeotropic composition)
Water-Nitric acid → maximum-boiling azeotrope (vapour richer in water below azeotropic composition)

Both have a point where composition is identical — but students miss that both are correct answers here.

How to avoid:

Learn the classic examples:

Azeotrope Type	Example	Composition same?
Minimum-boiling	Water-Ethanol (95.6% ethanol)	✓ Yes at that point
Maximum-boiling	Water-Nitric acid (68% HNO₃)	✓ Yes at that point
No azeotrope	Benzene-Toluene	✗ Never

Mistake 3: Thinking "same composition" means at all compositions

Why it's wrong:

The question asks: "will have same composition in liquid and vapour phase" — this happens only at the azeotropic point, not for all mixtures of those components.

How to avoid:

Read carefully: "will have" means there exists a composition where this is true. For azeotropic mixtures, yes. For ideal solutions, no such composition exists (except pure components, which aren't mixtures).

Mistake 4: Selecting only one option when multiple are correct

Why it's wrong:

The question explicitly says "[Note: two or more options may be correct]" — yet students often pick just (iii) Water-Ethanol, forgetting (ii) Water-Nitric acid.

How to avoid:

Always check the note at the top
For azeotropes: both minimum-boiling AND maximum-boiling azeotropes satisfy the condition
Memorise:
- Minimum-boiling azeotropes: Water-Ethanol, Water-Isopropanol, Acetone-Chloroform …

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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.[Note: two or more options may be correct] Which of the following binary mixtures will have same composition in liquid and vapour phase?

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