Explain why, on addition of 1 mol of KCl to 1 litre of water, the boiling point of water increases, while the

Question

Q. Explain why, on addition of 1 mol of KCl to 1 litre of water, the boiling point of water increases, while the addition of 1 mol of methyl alcohol to 1 litre of water decreases the boiling point.

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

🔒 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 — 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).

--- …

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

Boiling point elevation depends on the number of solute particles in solution. KCl dissociates into two ions (K⁺ and Cl⁻), doubling the particle count and raising the boiling point. Methyl alcohol does not dissociate — it is a volatile solute that escapes into vapour, lowering the vapour pressure of water less than expected, and in fact its own vapour adds to the total pressure, so the boiling point decreases.

The core idea: boiling point is about vapour pressure

A liquid boils when its vapour pressure equals the external atmospheric pressure. For pure water at 1 atm, that happens at 100 °C.

When you add a non‑volatile solute (like salt), the solute particles occupy space at the surface and reduce the number of water molecules that can escape into vapour. The vapour pressure of water drops. To make it reach 1 atm again, you must heat the solution to a higher temperature — that’s boiling point elevation.

But if the solute itself is volatile (like methyl alcohol), it also contributes its own vapour pressure. The total vapour pressure above the solution is the sum of water’s vapour pressure and the alcohol’s vapour pressure. That total can exceed 1 atm at a temperature below 100 °C, so the solution boils at a lower temperature.

The key difference: KCl is non‑volatile and dissociates; methyl alcohol is volatile and does not dissociate.

Step‑by‑step reasoning

1. What happens when KCl is added?

KCl is an ionic compound. In water it dissociates completely:

\text{KCl (s)} \rightarrow \text{K}^+ (aq) + \text{Cl}^- (aq)

So 1 mol of KCl produces 2 mol of ions in solution. Each ion acts as an independent solute particle.

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

\Delta T_b = i \cdot K_b \cdot m

where:

$i$ = van’t Hoff factor (number of particles per formula unit) — here $i = 2$
$K_b$ = ebullioscopic constant of water ( $0.512\ \text{°C·kg/mol}$ )
$m$ = molality (≈ 1 mol/kg for 1 mol in 1 L water, since 1 L water ≈ 1 kg)

Thus:

\Delta T_b = 2 \times 0.512 \times 1 = 1.024\ \text{°C}

The boiling point rises to about 101.0 °C.

Watch out

A common mistake is to forget the van’t Hoff factor. For KCl, $i = 2$ , not 1. If you used $i = 1$ , you’d get only 0.512 °C — still an elevation, but wrong in magnitude. The direction (increase) is correct either way, but the reason is the increased particle count.

2. What happens when methyl alcohol (CH₃OH) is added?

Methyl alcohol is a covalent, volatile liquid. It does not dissociate — it remains as single molecules. So $i = 1$ .

But here’s the crucial difference: methyl alcohol molecules themselves can escape into the vapour phase. The vapour above the solution contains both water and alcohol molecules.

Raoult’s law for a mixture of two volatile liquids says:

P_{\text{total}} = P_{\text{water}}^* \cdot x_{\text{water}} + P_{\text{alcohol}}^* \cdot x_{\text{alcohol}}

where $P^*$ is the vapour pressure of the pure component and $x$ is its mole fraction.

At 100 °C, $P_{\text{water}}^* = 1\ \text{atm}$ . For methyl alcohol, $P_{\text{alcohol}}^*$ at 100 °C is much higher — about 2.5 atm. Even a small mole fraction of alcohol adds significantly to the total pressure.

For 1 mol alcohol in 1 L water (≈ 55.5 mol water):

x_{\text{alcohol}} = \frac{1}{1 + 55.5} \approx 0.0177

x_{\text{water}} \approx 0.9823

Then:

P_{\text{total}} \approx (1 \times 0.9823) + (2.5 \times 0.0177) = 0.9823 + 0.0443 = 1.0266\ \text{atm}

That’s above 1 atm at 100 °C. So the solution will boil at a temperature below 100 °C — the boiling point decreases.

Tip

A quick way to see this: if the solute is more volatile than the solvent, the boiling point of the mixture lies between the boiling points of the two pure components. Pure methyl alcohol boils at 64.7 °C, so adding it to water pulls the boiling point downward.

3. Why the two cases are opposite

Property	KCl	Methyl alcohol
Nature	Ionic, non‑volatile	Covalent, volatile

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

Q.Explain why, on addition of 1 mol of KCl to 1 litre of water, the boiling point of water increases, while the addition of 1 mol of methyl alcohol to 1 litre of water decreases the boiling point.

The core idea: boiling point is about vapour pressure

Step‑by‑step reasoning

1. What happens when KCl is added?

2. What happens when methyl alcohol (CH₃OH) is added?

3. Why the two cases are opposite

🎓Unlock everything free for 14 days

Unlock everything free for 14 days