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

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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.

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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 key is the nature of the solute: NaCl dissociates into ions, increasing the number of particles and raising the boiling point (colligative effect), while methyl alcohol is volatile and lowers the vapour pressure of water less than its own vapour pressure adds, causing a net decrease in boiling point.

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.

\Delta T_b = i \cdot K_b \cdot m

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

\Delta T_b \approx 2 \times 0.512 \times 1 = 1.024\ \text{K}

The boiling point rises by about 1°C.

Watch out

A common mistake is to forget the van’t Hoff factor $i$ . NaCl gives two particles, so the effect is double that of a non-electrolyte like sugar. Always check if the solute dissociates.

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:

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

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.

Tip

Think of it this way: methanol is “eager” to escape into the vapour. Adding it to water makes the mixture’s vapour pressure higher than pure water’s at the same temperature. So the mixture reaches 1 atm at a lower temperature — the boiling point decreases.

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

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

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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:

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).
Apply Raoult's Law

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

P_{\text{solvent}} = X_{\text{solvent}} \cdot P_{\text{solvent}}^0

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

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:

P_{\text{total}} = X_{\text{water}}P_{\text{water}}^0 + X_{\text{alcohol}}P_{\text{alcohol}}^0

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

4. Relate to boiling point …

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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:

Property	Non-volatile solute	Volatile solute
Boiling point	Increases	Decreases
Freezing point	Decreases	Decreases

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: …

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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 21 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$K_f$
›Reveal solutionSolution
Using $\Delta T = K \times$ molality for both cases and $K_f = 4K_b$ , the ratio $Y:Z$ works out to $1:8$ .

Concept and Intuition

Both elevation of boiling point and depression of freezing point are colligative properties proportional to molality: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$ . Since the same solute A (molar mass 100) is used in the same mass of the same solvent X in both cases, only the mass of solute taken and the relevant constant ( $K_b$ vs $K_f$ ) differ between the two scenarios.

Step-by-Step Solution

Case 1 ( $\Delta T_b = Y$ ): moles of A $= \dfrac{2}{100} = 0.02$ mol; mass of solvent $= 0.2$ kg. Molality $m_1 = \dfrac{0.02}{0.2} = 0.1$ mol/kg.

$Y = K_b \times 0.1$

Case 2 ( $\Delta T_f = Z$ ): moles of A $= \dfrac{4}{100} = 0.04$ mol; mass of solvent $= 0.2$ kg. Molality $m_2 = \dfrac{0.04}{0.2} = 0.2$ mol/kg. $Z = K_f \times 0.2$ …
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AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$100.5°C$
›Reveal solutionSolution
This tests linking boiling-point elevation to freezing-point depression via a common molality; the freezing point works out to $-1.816°C$ .

Concept and Intuition

For a dilute solution of a non-volatile, non-electrolyte solute, both the elevation in boiling point and the depression in freezing point are colligative properties proportional to the same molality of the solute: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$ . Given one property, we can back out the molality and then predict the other.

Step-by-Step Solution

Elevation in boiling point: $\Delta T_b = 100.5°C - 100°C = 0.5\ K$ .

Find molality from $\Delta T_b = K_b m$ : $m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.5}{0.512} \approx 0.9766\ mol\,kg^{-1}$ .

Find depression in freezing point: $\Delta T_f = K_f m = 1.86 \times 0.9766 \approx 1.8164\ K$ . …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$KCl$
›Reveal solutionSolution
This tests boiling-point elevation with a dissociating electrolyte, requiring first converting molarity to molality via the given density. The boiling point comes out to 100.996 °C.

Concept and Intuition

Boiling point elevation depends on molality (not molarity), and for an electrolyte that partially dissociates, the van't Hoff factor $i$ accounts for the actual number of particles in solution.

Step-by-Step Solution

Take 1 litre (1000 mL) of the 1M KCl solution as the basis. Moles of KCl $= 1$ mol.

Mass of the whole solution $= \text{density} \times \text{volume} = 1.04\ g/mL \times 1000\ mL = 1040\ g$ .

Mass of solute (KCl) $= 1\ mol \times 74.5\ g/mol = 74.5\ g$ .

Mass of solvent (water) $= 1040 - 74.5 = 965.5\ g = 0.9655\ kg$ .

Molality $m = \dfrac{\text{moles solute}}{\text{kg solvent}} = \dfrac{1}{0.9655} \approx 1.0357\ mol\,kg^{-1}$ .

KCl dissociates as $KCl \to K^+ + Cl^-$ , so $n = 2$ ions per formula unit. With degree of dissociation $\alpha = 0.85$ : van't Hoff factor $i = 1 + \alpha(n-1) = 1 + 0.85(2-1) = 1.85$ . …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$x$
›Reveal solutionSolution
Using the osmotic-pressure data to find the molar concentration, then the molality, and finally applying $\Delta T_b = iK_bm$ gives $\Delta T_b \approx 0.016\ K$ .

Concept and Intuition

Both osmotic pressure and boiling-point elevation are colligative properties that depend on the effective number of particles in solution (captured by the van't Hoff factor $i$ ). We first use the osmotic pressure relation $\pi = iCRT$ to back out the molar concentration $C$ of $CaCl_2$ , convert that to molality using the mass of water given, and then apply the boiling-point elevation formula.

Step-by-Step Solution

From $\pi = iCRT$ : $C = \dfrac{\pi}{iRT} = \dfrac{0.75}{2.5 \times 0.08 \times 300} = \dfrac{0.75}{60} = 0.0125\ mol\ L^{-1}$ .

Moles of $CaCl_2$ , $x = C \times V = 0.0125 \times 2.5\ L = 0.03125\ mol$ .

Mass of water (solvent) $= 2.5\ L \times 1000\ g/L\ (\text{density}=1\ g/mL) = 2500\ g = 2.5\ kg$ .

Molality, $m = \dfrac{x}{\text{kg of solvent}} = \dfrac{0.03125}{2.5} = 0.0125\ mol\ kg^{-1}$ . …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$K_f$
›Reveal solutionSolution
This links two colligative properties (boiling point elevation and freezing point depression) through the common molality of the solution.

Concept and Intuition

Both boiling-point elevation and freezing-point depression are colligative properties proportional to the same molality of solute particles: $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$ . Given one, we can find the molality and then use it to get the other.

Step-by-Step Solution

Boiling point elevation: $\Delta T_b = 100.18 - 100 = 0.18\,^\circ C$ .

Molality from $\Delta T_b = K_b m$ : $m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.18}{0.52} = 0.3462\ \text{mol/kg}$ .

Freezing point depression: $\Delta T_f = K_f m = 1.86 \times 0.3462 = 0.6439\,^\circ C \approx 0.64\,^\circ C$ . …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$K_f$
›Reveal solutionSolution
The cryoscopic-constant formula relates $K_f$ to the solvent's molar mass, freezing point, and molar enthalpy of fusion; solving gives $\Delta_{fus}H \approx 6$ kJ/mol for water.

Concept and Intuition

The molal depression constant $K_f$ of a solvent is fundamentally derived from its freezing point and its enthalpy of fusion (the energy needed to melt the solid), via the thermodynamic relation:
$K_f = \dfrac{R\,T_f^2\,M}{1000\,\Delta_{fus}H}$
where $M$ is the solvent's molar mass (g/mol), $T_f$ its freezing point (K), and $\Delta_{fus}H$ its molar enthalpy of fusion (J/mol).

Step-by-Step Solution

Rearranging for $\Delta_{fus}H$ : $\Delta_{fus}H = \dfrac{R\,T_f^2\,M}{1000\,K_f}$ .

Substitute $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , $T_f=273$ K, $M=18\ \mathrm{g/mol}$ , $K_f=1.86\ \mathrm{K\,kg\,mol^{-1}}$ .

$T_f^2 = 74529$ . …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$\Delta T_b$
›Reveal solutionSolution
From the boiling-point elevation, the molality is found (0.1 mol/kg), and then the SAME molality is used with Kf to get the freezing-point depression, giving a freezing point of 272.814 K.

Concept and Intuition

Both boiling-point elevation and freezing-point depression are colligative properties governed by the same solution molality — ΔTb = Kb·m and ΔTf = Kf·m. Since molality doesn't change, we can find it from one property and use it to compute the other.

Step-by-Step Solution

ΔTb = Kb·m → m = ΔTb/Kb = 0.052 / 0.52 = 0.1 mol/kg.

ΔTf = Kf·m = 1.86 × 0.1 = 0.186 K.

Freezing point of solution = freezing point of pure water − ΔTf = 273 − 0.186 = 272.814 K. …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$T_1, T_2, T_3, T_4$
›Reveal solutionSolution
The solution's boiling point is where its vapour-pressure curve meets $P=1$ atm, which is $T_3$ on the graph (higher than pure water's $T_2$ , consistent with boiling point elevation).

Concept and Intuition

A liquid boils at the temperature where its vapour pressure equals the surrounding atmospheric pressure. Dissolving a non-volatile solute (urea) lowers the solution's vapour pressure at any given temperature (Raoult's law), which means the solution's vapour-pressure curve sits to the right of pure water's curve — it needs a HIGHER temperature to reach the same vapour pressure. This is exactly boiling point elevation, $\Delta T_b > 0$ .

Step-by-Step Solution

Atmospheric pressure is given as 1 atm, so the boiling point of any liquid on this graph is where its curve crosses the horizontal $y=1.00$ line.

The problem states the left curve (pure water, lower vapour pressure needed at lower temperature) crosses $y=1.00$ at $T_2$ .

The right curve (urea solution, shifted to higher temperature for the same vapour pressure due to boiling point elevation) crosses $y=1.00$ at $T_3$ . …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\Delta T_f$
›Reveal solutionSolution
The same molality drives both boiling-point elevation and freezing-point depression; find $m$ from $\Delta T_b$ , then use it with $K_f$ to get $\Delta T_f$ .

Concept and Intuition

Both colligative properties depend on the same solution molality $m$ : $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$ . Since urea is a non-electrolyte (no dissociation), the molality computed from one property applies directly to the other.

Step-by-Step Solution

$m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.104}{0.52} = 0.2\ \text{mol/kg}$ . …
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AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$K_b$
›Reveal solutionSolution
Since KCl fully dissociates into 2 ions per formula unit (van't Hoff factor $i=2$ ), the boiling point elevation is $\Delta T_b = i K_b m = 0.104$ K, giving a boiling point of 373.104 K.

Concept and Intuition

Boiling point elevation is a colligative property that depends on the total number of solute particles in solution, not just the number of formula units dissolved. For an electrolyte like KCl that dissociates completely into K $^+$ and Cl $^-$ ( $\alpha = 100\%$ ), each mole of KCl produces 2 moles of particles, so the van't Hoff factor $i = 2$ must be included in the elevation formula.

Step-by-Step Solution

Formula: $\Delta T_b = i \, K_b \, m$ .

Since KCl dissociates completely ( $\alpha=100\%$ ) into 2 ions (K $^+$ + Cl $^-$ ), $i = 1 + \alpha(n-1) = 1 + 1\times(2-1) = 2$ .

Substitute: $\Delta T_b = 2 \times 0.52 \times 0.1 = 0.104$ K. …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$K_f$
›Reveal solutionSolution
Converting the given mole fraction of solute to molality and applying $\Delta T_f=K_f m$ gives a freezing-point depression of about 1.043 K.

Concept and Intuition

Freezing point depression depends on molality, not mole fraction directly, so we must first convert. For a solution with total 1 mole (basis), if $x_2=0.01$ is the solute's mole fraction, then $n_2=0.01$ and $n_1=0.99$ (moles of water). Molality is moles of solute per kg of solvent.

Step-by-Step Solution

Take a basis of 1 total mole: $n_2 = 0.01$ mol solute, $n_1 = 0.99$ mol water.

Mass of water (solvent): $0.99\ \text{mol}\times18\ \text{g/mol} = 17.82$ g $=0.01782$ kg.

Molality $m = \dfrac{n_2}{\text{mass of solvent in kg}} = \dfrac{0.01}{0.01782} \approx 0.5612$ mol/kg. …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\mathrm{mol^{-1}}$
›Reveal solutionSolution
Freezing-point depression gives the molality directly; converting molality to mass via the given molar mass gives $x \approx 20.96\,g$ .

Concept and Intuition

Freezing-point depression is a colligative property: $\Delta T_f = K_f \times m$ , where $m$ is the molality of the solute (moles of solute per kg of solvent) and $K_f$ is the cryoscopic constant of the solvent. Since the solute is non-volatile and (implicitly) a non-electrolyte (no van't Hoff factor mentioned), we use this formula directly.

Step-by-Step Solution

Given: $\Delta T_f = 1.0^{\circ}C$ , $K_f = 1.86\ K\,kg\,mol^{-1}$ , mass of water $=0.5\ kg$ , molar mass of solute $M = 78\ g\,mol^{-1}$ .

Find molality: $m = \dfrac{\Delta T_f}{K_f} = \dfrac{1.0}{1.86} = 0.5376\ mol\,kg^{-1}$ .

Molality is moles of solute per kg solvent, so moles of solute $n = m \times (\text{mass of water in kg}) = 0.5376 \times 0.5 = 0.2688\ mol$ . …
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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.

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

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

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