Assertion: When methyl alcohol is added to water, boiling point of water increases. Reason: When a volatile so

NCERT Exemplar · Q52

Q. Assertion: When methyl alcohol is added to water, boiling point of water increases. Reason: When a volatile solute is added to a volatile solvent elevation in boiling point is observed. Choose the correct option:

(i) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(iii) Assertion is correct statement but reason is wrong statement.

(iv) Assertion and reason both are incorrect statements.

(v) Assertion is wrong statement but reason is correct statement.

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Concept understanding — Boiling Point Elevation

Boiling Point Elevation — From Intuition to Precision

Imagine you're cooking pasta. You add salt to the water, and the water takes longer to boil. That's not your imagination — it's a real physical effect. The salt has raised the boiling point of the water. Pure water boils at 100°C at sea level, but salt water needs a slightly higher temperature to boil. That's boiling point elevation in action.

Why does this happen?

To understand why, you need to think about what boiling actually is. Boiling occurs when the vapour pressure of the liquid equals the surrounding atmospheric pressure. Vapour pressure is the pressure exerted by molecules escaping from the liquid surface into the gas phase.

When you dissolve a non-volatile solute (like salt, which doesn't evaporate) in a solvent (like water), something interesting happens at the surface. Some of the surface molecules are now solute particles — they don't escape into the vapour. This means fewer solvent molecules can leave the liquid per second. The result? The vapour pressure of the solution is lower than that of the pure solvent at the same temperature.

Now, to make this reduced vapour pressure equal to the atmospheric pressure (so the liquid can boil), you need to raise the temperature. That extra heat gives the remaining solvent molecules enough energy to overcome the solute's interference and produce the required vapour pressure.

Note

The solute itself doesn't boil away — it stays behind. That's why we call it a non-volatile solute. If the solute were volatile (like alcohol), the story changes.

The precise statement

Boiling point elevation is the increase in the boiling point of a solvent when a non-volatile solute is dissolved in it. The boiling point of the solution is always higher than that of the pure solvent.

The elevation $\Delta T_b$ is given by:

\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})

And it depends only on the number of solute particles, not on their chemical identity (for dilute solutions). This is a colligative property — a property that depends on the concentration of particles, not their nature.

\Delta T_b = K_b \cdot m

What is $K_b$ ?

The ebullioscopic constant $K_b$ is a property of the pure solvent. It tells you how much the boiling point rises per unit molal concentration. For water, $K_b = 0.512 \, \text{°C kg mol}^{-1}$ . So a 1 molal aqueous solution (1 mole of solute per kg of water) boils at $100.512\,°\text{C}$ .

Tip

For quick calculations: $\Delta T_b \propto m$ . Double the molality, double the elevation. But this works only for dilute solutions — at high concentrations, interactions between solute particles break the linearity.

What about electrolytes?

If the solute dissociates into ions (like NaCl → Na⁺ + Cl⁻), the number of particles increases. One mole of NaCl gives two moles of particles in solution. So the effective concentration is higher. We account for this using the van't Hoff factor $i$ :

\Delta T_b = i \cdot K_b \cdot m

For NaCl, $i \approx 2$ (in dilute solutions). For sugar (which doesn't dissociate), $i = 1$ . …

Why this formula?

Boiling Point Elevation: Why the Formula Holds

Let's build this from first principles — understanding the why before the what.

What Happens at the Boiling Point?

At the boiling point, a liquid's vapor pressure equals the external atmospheric pressure. For a pure solvent, this happens at a fixed temperature (e.g., 100°C for water at 1 atm).

When you add a non-volatile solute (like salt or sugar), the solute particles occupy space at the liquid surface, reducing the number of solvent molecules that can escape into vapor. This lowers the vapor pressure of the solution compared to the pure solvent.

The Key Consequence

Since vapor pressure is now lower, the solution must be heated to a higher temperature to make its vapor pressure reach atmospheric pressure again. That's the boiling point elevation.

The Formula and Its Derivation

The boiling point elevation $\Delta T_b$ is given by:

\Delta T_b = K_b \cdot m

where:

$\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})$
$K_b$ = ebullioscopic constant (depends only on the solvent)
$m$ = molality of the solution (moles of solute per kg of solvent)

Why Molality and Not Molarity?

Molality is temperature-independent — it doesn't change when the solution is heated. Molarity (moles per liter) changes because volume expands with temperature. Since we're measuring a temperature change, we need a concentration unit that stays fixed.

The Reasoning Behind $K_b$

The constant $K_b$ comes from thermodynamics. For a dilute solution, the vapor pressure lowering follows Raoult's Law:

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

where $x_{\text{solvent}}$ is the mole fraction of solvent and $P_{\text{solvent}}^0$ is the pure solvent's vapor pressure.

Using the Clausius-Clapeyron equation (which relates vapor pressure to temperature) and Raoult's Law, one can derive:

K_b = \frac{R T_b^2}{1000 \cdot \Delta H_{\text{vap}}}

where:

$R$ = gas constant (8.314 J/mol·K)
$T_b$ = boiling point of pure solvent (in Kelvin)
$\Delta H_{\text{vap}}$ = molar enthalpy of vaporization (J/mol)
The factor 1000 converts grams to kg (since molality uses kg of solvent)

What This Tells Us

$K_b$ is a property of the solvent alone — it doesn't depend on the solute.
Solvents with higher boiling points or lower heats of vaporization have larger $K_b$ values (greater elevation per molal concentration).

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The assertion is wrong (adding methyl alcohol to water lowers the boiling point), and the reason is also wrong (a volatile solute in a volatile solvent does not show the usual boiling point elevation). So both statements are incorrect.

The core idea: Henry’s Law and vapour pressure

Boiling happens when the vapour pressure of the liquid equals the external (atmospheric) pressure.

Adding a non-volatile solute (like salt or sugar) to a solvent lowers the solvent’s vapour pressure — so you need a higher temperature to reach atmospheric pressure. That’s the familiar elevation in boiling point.

But here, methyl alcohol (methanol) is volatile — it has its own significant vapour pressure. When you add a volatile solute to a volatile solvent, the total vapour pressure above the mixture is the sum of the partial pressures of both components (Raoult’s law for ideal mixtures). That sum can be higher than the vapour pressure of pure water, not lower.

P_{\text{total}} = P_A^\circ x_A + P_B^\circ x_B

Step-by-step reasoning

What happens to the boiling point when you add methanol to water?

Methanol has a much higher vapour pressure than water at any given temperature (it boils at $65^\circ\text{C}$ , water at $100^\circ\text{C}$ ). Adding methanol to water increases the total vapour pressure of the mixture above that of pure water.

Since the mixture now reaches atmospheric pressure at a lower temperature, the boiling point decreases.

So the assertion — “boiling point of water increases” — is false.
Is the reason statement correct?

The reason says: “When a volatile solute is added to a volatile solvent, elevation in boiling point is observed.” …

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Here’s a breakdown of the common mistakes students make with this question, and how to avoid each.

The Core Concept: Colligative Properties & Volatile Solutes

This question tests your understanding of colligative properties — properties that depend on the number of solute particles, not their identity. The key twist here is that both the solute and solvent are volatile.

Normal boiling point elevation: Adding a non-volatile solute (like salt) to a volatile solvent (like water) raises the boiling point. The solute particles block solvent molecules from escaping, so you need a higher temperature to boil.
Volatile solute + volatile solvent: If the solute is also volatile (like methyl alcohol), it lowers the partial pressure of the solvent, but it adds its own vapor pressure. The net effect on the boiling point depends on the relative volatilities. In the case of methyl alcohol (more volatile than water) added to water, the boiling point of the solution decreases (or remains nearly the same, but definitely does not increase).

Common Mistake #1: Assuming all solutes raise the boiling point

The Mistake: Students blindly apply the rule “adding a solute raises the boiling point” without checking if the solute is volatile. They assume the Assertion is correct.

Why it happens: The standard textbook example is salt in water. Students memorize the result without understanding the mechanism (vapor pressure lowering).

How to Avoid:

Always ask: “Is the solute volatile?” If yes, the colligative property rule for boiling point elevation does not apply in the same way.
Remember the mechanism: Boiling point elevation happens because the solute reduces the vapor pressure of the solvent. A volatile solute adds its own vapor pressure, which can increase the total vapor pressure, thus lowering the boiling point.
Key fact: Methyl alcohol (methanol) is more volatile than water. Adding it to water makes the solution easier to boil (lower boiling point).

Correct understanding: The Assertion is wrong. The boiling point of water decreases when methyl alcohol is added.

Common Mistake #2: Confusing “volatile” with “non-volatile” in the Reason

The Mistake: Students think the Reason is correct because it sounds like a standard textbook statement. They don’t notice the critical word “volatile” in the Reason.

Why it happens: The Reason says: “When a volatile solute is added to a volatile solvent elevation in boiling point is observed.” This is a false statement. Students often read it quickly and mentally replace “volatile” with “non-volatile.”

How to Avoid:

Read every word carefully. The Reason explicitly says “volatile solute” and “volatile solvent.” That combination does not give boiling point elevation.
Create a mental checklist:
- Non-volatile solute + volatile solvent → Elevation in boiling point.
- Volatile solute + volatile solvent → Depression or no change in boiling point (depends on relative volatilities).
Practice with examples: Sugar (non-volatile) in water → elevation. Alcohol (volatile) in water → depression.

Correct understanding: The Reason is wrong because it states the opposite of the correct behavior.

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

TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$_2$
›Reveal solutionSolution
The key idea is that the temperature of maximum density for water isotopes shifts due to isotopic mass differences; D₂O has its maximum density at about 11.2 °C (284.35 K) and H₂O at 3.98 °C (277.13 K), so the difference is roughly 7.2 K, making option (A) 7.0 the closest.

Concept and intuition

Most liquids become denser as they cool, but water is famously anomalous: it reaches its maximum density at about 4 °C, not at its freezing point. This happens because of the balance between thermal contraction and the open, hydrogen-bonded structure that forms as water approaches freezing. For heavy water (D₂O), the stronger deuterium bonds shift this balance to a higher temperature. The difference in these maximum-density temperatures is what we need.

Step-by-step reasoning

Recall the known values

For ordinary water (H₂O), the temperature of maximum density is well known:

$T_{\text{max}}(\text{H}_2\text{O}) = 3.98^\circ\text{C} \approx 4^\circ\text{C}$
In Kelvin:
$y = 273.15 + 3.98 = 277.13\ \text{K}$

Find the corresponding value for heavy water (D₂O)

Heavy water’s maximum density occurs at a higher temperature because D₂O forms stronger hydrogen bonds (due to the greater mass of deuterium), so the open structure persists to higher temperatures.

The accepted experimental value is:

$T_{\text{max}}(\text{D}_2\text{O}) \approx 11.2^\circ\text{C}$
In Kelvin: $$ x = 273.15 + 11.2 = 284.35\ \text{K} …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$_2$
›Reveal solutionSolution
The key idea is that the temperature of maximum density for water isotopes shifts due to isotopic mass differences; D₂O has its maximum density at about 11.2 °C (284.35 K) and H₂O at 3.98 °C (277.13 K), so the difference is roughly 7.2 K, making option (B) 7.0 the closest.

Why this approach works

The temperature of maximum density of water is a well-known physical property: for ordinary H₂O it is 3.98 °C. For heavy water (D₂O), the stronger hydrogen bonds (due to the greater mass of deuterium) shift the density maximum to a higher temperature. The difference arises from isotopic effects on molecular vibrations and hydrogen-bond strength. Instead of deriving from first principles, we recall the accepted experimental values: D₂O’s density maximum occurs near 11.2 °C. Converting to Kelvin and subtracting gives the answer.

Step-by-step reasoning

Recall the known value for H₂O The temperature of maximum density for ordinary water (H₂O) is $3.98\,^\circ\text{C}$ . In Kelvin:

$y = 3.98 + 273.15 = 277.13\ \text{K}$

Recall the known value for D₂O Heavy water (D₂O) has its maximum density at about $11.2\,^\circ\text{C}$ (the exact value is often cited as 11.2 °C or 11.23 °C). In Kelvin:

$x = 11.2 + 273.15 = 284.35\ \text{K}$

Compute the difference
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$0.1\,\text{M}\ \mathrm{BaCl}_2$
›Reveal solutionSolution
Colligative properties depend on the number of particles in solution, not their identity. The substance that dissociates into the most ions will show the highest colligative effect. Here, $0.1\,\text{M}\ (\mathrm{NH}_4)_3\mathrm{PO}_4$ gives the most particles (4 ions per formula unit), so it wins.

Colligative properties — like boiling point elevation, freezing point depression, and osmotic pressure — depend only on the number of solute particles in a given amount of solvent. That’s the core idea. So when you compare equimolar solutions (all at $0.1\,\text{M}$ ), the one that breaks into the most ions in water will have the highest effective particle concentration, and therefore the strongest colligative effect.

Let’s check each option.

$0.1\,\text{M}\ \mathrm{BaCl}_2$ Barium chloride dissociates fully in water:

$\mathrm{BaCl}_2 \rightarrow \mathrm{Ba}^{2+} + 2\mathrm{Cl}^-$
That’s 3 ions per formula unit. So the total particle concentration is $0.1 \times 3 = 0.3\,\text{M}$ .

$0.1\,\text{M}\ \mathrm{AgNO}_3$ Silver nitrate dissociates:

$\mathrm{AgNO}_3 \rightarrow \mathrm{Ag}^+ + \mathrm{NO}_3^-$
That’s 2 ions per formula unit. Particle concentration: $0.1 \times 2 = 0.2\,\text{M}$ .

$0.1\,\text{M}\ \text{urea}$

Urea is a covalent, non-electrolyte. It does not dissociate at all. So it remains as 1 particle per molecule. Particle concentration: $0.1\,\text{M}$ .

$0.1\,\text{M}\ (\mathrm{NH}_4)_3\mathrm{PO}_4$

Ammonium phosphate dissociates: …
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TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$T_f$
›Reveal solutionSolution
Freezing point depression obeys $\Delta T_f = K_f m$ , so the freezing point itself falls linearly with molality: $T_f = T_f^{\circ} - K_f m$ . The graph is a straight line with negative slope — option (B).

The concept first: why does a solute lower the freezing point?

A liquid freezes at the temperature where the vapour pressure of the liquid equals the vapour pressure of the solid.

Dissolve a non-volatile solute in the solvent. The solute particles occupy part of the surface, so fewer solvent molecules escape — the vapour pressure of the solution falls (Raoult's law).

The vapour pressure of the pure solid solvent is unaffected.

So the two curves — solution and solid — now intersect at a lower temperature. That new intersection is the solution's freezing point.

Hence: the freezing point of a solution is always lower than that of the pure solvent, and the more solute you add, the lower it goes.

Because the effect depends only on the number of solute particles, not on their identity, freezing-point depression is a colligative property.

Step-by-step derivation of the graph

Experiment and thermodynamics both give

$\Delta T_f \;\propto\; m \qquad\Longrightarrow\qquad \Delta T_f = K_f\, m$
where $m$ is the molality and $K_f$ the cryoscopic constant (a fixed property of the solvent — for water, $K_f = 1.86\ \text{K kg mol}^{-1}$ ).

By definition,

$\Delta T_f = T_f^{\circ} - T_f$
where $T_f^{\circ}$ is the freezing point of the pure solvent and $T_f$ that of the solution.

Substitute:

$T_f^{\circ} - T_f = K_f\, m$ $\boxed{\,T_f = T_f^{\circ} - K_f\, m\,}$

Compare with the equation of a straight line, $y = c + (\text{slope})x$ :

Variable on the $y$ -axis: $T_f$

Variable on the $x$ -axis: $m$

Intercept at $m = 0$ : $T_f^{\circ}$ — the pure solvent's freezing point (a positive, finite value)

Slope: $-K_f$ — a negative constant …
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Q.Assertion: When methyl alcohol is added to water, boiling point of water increases. Reason: When a volatile solute is added to a volatile solvent elevation in boiling point is observed. Choose the correct option:

Concept understanding — Boiling Point Elevation

Boiling Point Elevation — From Intuition to Precision

Why does this happen?

The precise statement

What is $K_b$ ?

What about electrolytes?

Why this formula?

Boiling Point Elevation: Why the Formula Holds

What Happens at the Boiling Point?

The Key Consequence

The Formula and Its Derivation

Why Molality and Not Molarity?

The Reasoning Behind $K_b$

What This Tells Us

The core idea: Henry’s Law and vapour pressure

Step-by-step reasoning

Concept: Colligative Properties & Volatile Solutes

Method: Conceptual Analysis of Boiling Point Elevation with Volatile Solute

Step 1: Understand the Assertion

Step 2: Understand the Reason

The Core Concept: Colligative Properties & Volatile Solutes

Common Mistake #1: Assuming all solutes raise the boiling point

Common Mistake #2: Confusing “volatile” with “non-volatile” in the Reason

Why this approach works

Step-by-step reasoning

The concept first: why does a solute lower the freezing point?

Step-by-step derivation of the graph

Q.Assertion: When methyl alcohol is added to water, boiling point of water increases. Reason: When a volatile solute is added to a volatile solvent elevation in boiling point is observed. Choose the correct option:

The core idea: Henry’s Law and vapour pressure

Step-by-step reasoning

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