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

The key idea is boiling point elevation for non-volatile solutes versus boiling point depression when a volatile solute is added.

Reasoning:

1. NaCl (non-volatile solute): NaCl dissociates into ions ($\text{Na}^+$ and $\text{Cl}^-$), increasing the number of particles in solution. This lowers the vapour pressure of water. To reach the external pressure (1 atm), the solution must be heated to a higher temperature — hence the boiling point rises. …

When you add a solute to a solvent, the boiling point can either rise or fall depending on whether the solute is non-volatile or volatile. This is not a contradiction — it’s a direct consequence of Raoult’s law and the definition of boiling point.

Boiling occurs when the vapour pressure of the liquid equals the external atmospheric pressure. Adding a non-volatile solute lowers the vapour pressure of the solvent, so you need a higher temperature to reach atmospheric pressure — hence boiling point elevation. But if the solute itself is volatile, it contributes its own vapour pressure, and the total vapour pressure may become higher than that of the pure solvent at the same temperature. In that case, the boiling point can actually drop.

Let’s examine each case.

1. NaCl in water — a non-volatile, ionic solute

NaCl dissolves completely into Na⁺ and Cl⁻ ions. Each mole of NaCl gives 2 moles of particles in solution (assuming complete dissociation). Water itself has a negligible vapour pressure at its boiling point, and the ions are non-volatile — they do not escape into the vapour phase.

The vapour pressure of the solution is therefore lower than that of pure water at any given temperature. To raise this reduced vapour pressure up to 1 atm (the external pressure), you must heat the solution to a higher temperature. This is the classic boiling point elevation.

For 1 mol of NaCl in 1 L of water (≈ 1 kg water), $m \approx 1\ \text{mol/kg}$, so

The boiling point rises by about 1°C.

2. Methyl alcohol (CH₃OH) in water — a volatile solute

Methyl alcohol is volatile — it has a significant vapour pressure at room temperature and even more so near the boiling point of water. When you add 1 mol of methanol to 1 L of water, you create a binary liquid mixture where both components contribute to the total vapour pressure.

According to Raoult’s law for an ideal solution (and methanol–water is nearly ideal), the total vapour pressure above the solution is:

where $P^*$ are the vapour pressures of the pure components at a given temperature, and $x$ are mole fractions.

At the boiling point of pure water (100°C), $P_{\text{water}}^* = 1\ \text{atm}$. But methanol boils at only 64.7°C, so at 100°C its vapour pressure is much higher than 1 atm. Even a small mole fraction of methanol adds substantially to the total vapour pressure.

Let’s estimate: For 1 mol methanol in about 55.5 mol water (1 L ≈ 55.5 mol),

$x_{\text{methanol}} \approx \frac{1}{56.5} \approx 0.0177$ …

Concept: Colligative Properties & Vapour Pressure

The boiling point of a liquid depends on its vapour pressure. A liquid boils when its vapour pressure equals the atmospheric pressure. Any solute that changes the vapour pressure will change the boiling point.

Method: Raoult's Law & Boiling Point Elevation/Depression

Name of method: Raoult's Law analysis for non-volatile vs volatile solutes.

Steps:

1. Identify solute type

   • NaCl is a non-volatile ionic solute (it does not evaporate).
   • Methyl alcohol (CH₃OH) is a volatile molecular solute (it evaporates readily).

2. Apply Raoult's Law

   For a solution, the partial vapour pressure of a component is:

where $X$ is mole fraction and $P^0$ is vapour pressure of pure solvent.

3. Effect on total vapour pressure
   • NaCl case: Non-volatile solute reduces the mole fraction of water ($X_{\text{water}} < 1$). Since NaCl contributes zero vapour pressure, total vapour pressure decreases.
   • Methyl alcohol case: Both water and methyl alcohol are volatile. Adding CH₃OH introduces a second volatile component with its own vapour pressure. The total vapour pressure becomes:

 Since methyl alcohol has a **higher vapour pressure** than water at the same temperature, the total vapour pressure **increases**.

4. Relate to boiling point …

Here’s a clear breakdown of the common mistakes students make on this concept, along with how to avoid each.

The Core Concept

This question tests your understanding of colligative properties — specifically, boiling point elevation and vapour pressure.

• NaCl is a non-volatile solute. It does not evaporate. Adding it lowers the vapour pressure of water, so you need a higher temperature to reach atmospheric pressure → boiling point increases.
• Methyl alcohol (methanol) is a volatile solute. It evaporates easily and has its own vapour pressure. Adding it to water increases the total vapour pressure of the solution → the mixture boils at a lower temperature than pure water.

Common Mistake #1: Treating both solutes as identical

The error:

Students assume that since both are solutes, both should raise the boiling point. They apply the formula $\Delta T_b = i \cdot K_b \cdot m$ blindly, without checking whether the solute is volatile.

Why it happens:

Most textbook problems only deal with non-volatile solutes (like NaCl, sugar). Students memorise the formula but forget the condition: it only applies to non-volatile solutes.

How to avoid:

Always ask: “Does this solute evaporate?”

• If non-volatile → boiling point increases (e.g., NaCl, sugar).
• If volatile → boiling point decreases (e.g., methanol, ethanol, acetone).

Remember: The boiling point elevation formula $\Delta T_b = i K_b m$ is only valid for non-volatile solutes.

Common Mistake #2: Confusing boiling point with freezing point

The error:

Students mix up the two concepts — they think “adding anything to water always raises the boiling point and lowers the freezing point.” This is only half true.

Why it happens:

Both are colligative properties, but the volatility of the solute affects boiling point differently. Freezing point depression works for all solutes (volatile or not), but boiling point does not.

How to avoid:

Make a mental table:

Key distinction: Freezing point always drops. Boiling point only rises if the solute doesn’t evaporate.

Common Mistake #3: Ignoring the role of vapour pressure

The error:

Students give answers like “NaCl increases boiling point because it’s a salt” without explaining the mechanism — vapour pressure lowering.

Why it happens:

They memorise the result but not the reasoning. In exams, this loses marks for incomplete explanation.

How to avoid:

Always explain using vapour pressure:

• NaCl case:

  Water molecules at the surface are partially replaced by Na⁺ and Cl⁻ ions. Fewer water molecules escape into vapour → vapour pressure decreases → higher temperature needed to reach 1 atm → boiling point rises.

• Methanol case:

  Methanol molecules are volatile. They escape easily and add to the vapour. The total vapour pressure (water + methanol) is higher than pure water’s vapour pressure → the solution reaches 1 atm at a lower temperature → boiling point falls.

Formula to remember: …