(a) Calculate the elevation of boiling point of a solution when 3 g of $CaCl_2$ (Molar mass = 111 g mol$^{-1}$) was dissolved in 260 g of water, assuming that $CaCl_2$ undergoes complete dissociation. ($K_b$ for water = 0·52 K kg mol$^{-1}$) OR (b) Liquids 'X' and 'Y' form an ideal solution. The vapour pressure of pure 'X' and pure 'Y' are 120 mm Hg and 160 mm Hg respectively. Calculate the vapour pressure of the solution containing equal moles of 'X' and 'Y'.

> [!TLDR] > Part (a): $\Delta T_b = i K_b m = 3\times0.52\times0.104 \approx 0.16\ \text{K}$. > Part (b): By Raoult's law $P = 0.5(120)+0.5(160) = 140\ \text{mm Hg}$. ## Part (a) Boiling-point elevation is a colligative property that depends on the number of solute particles. For an electrolyte we include the van't Hoff factor $i$: > [!FORMULA] > $$\Delta T_b = i\,K_b\,m$$ **Van't Hoff factor:** $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$ produces $1+2 = 3$ ions, so with complete dissociation $i = 3$. **Molality:** $$\text{moles } CaCl_2 = \frac{3}{111}=0.0270\ \text{mol}, \quad \text{water}=0.260\ \text{kg}$$ $$m = \frac{0.0270}{0.260}=0.1040\ \text{mol kg}^{-1}$$ **Elevation:** …

(a) Calculate the elevation of boiling point of a solution when 3 g of CaCl₂ (Molar mass = 111 g mol⁻¹) was di

Question

CaCl_2

(OR)

(b) Liquids 'X' and 'Y' form an ideal solution. The vapour pressure of pure 'X' and pure 'Y' are 120 mm Hg and 160 mm Hg respectively. Calculate the vapour pressure of the solution containing equal moles of 'X' and 'Y'.

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Part (a)Concept understanding — Elevation of Boiling Point

Why does salt make water boil hotter?

You already know that pure water boils at $100^\circ\text{C}$ at 1 atm pressure. But if you dissolve salt or sugar in that water, the boiling point goes up. Not by much — a pinch of salt raises it by a fraction of a degree — but measurably. The question is: why?

Think about what boiling actually is. A liquid boils when its vapour pressure equals the surrounding atmospheric pressure. At that point, bubbles of vapour can form anywhere inside the liquid, not just at the surface. So boiling temperature is really the temperature at which the liquid's vapour pressure hits the external pressure.

Now add a non-volatile solute — something like salt or sugar that does not itself evaporate. The solute particles stay behind in the liquid. They get in the way of solvent molecules trying to escape into the vapour phase. Fewer solvent molecules make it to the surface per second, so the vapour pressure of the solution is lower than that of the pure solvent at the same temperature.

If the vapour pressure is lower, you need to heat the solution to a higher temperature to bring that vapour pressure back up to atmospheric pressure. That higher temperature is the new boiling point. The difference between this new boiling point and the pure solvent's boiling point is the elevation of boiling point, denoted $\Delta T_b$ .

Note

The solute must be non-volatile. If the solute itself evaporates (like alcohol in water), the reasoning changes completely — both components contribute to vapour pressure.

The precise statement

For dilute solutions of a non-volatile solute, the elevation of boiling point is directly proportional to the molal concentration of the solute. Molality ( $m$ ) is the number of moles of solute per kilogram of solvent.

\Delta T_b \propto m

Introducing the proportionality constant $K_b$ , called the ebullioscopic constant (or boiling point elevation constant) of the solvent:

\Delta T_b = K_b \cdot m

Here:

$\Delta T_b$ is the boiling point elevation (in K or °C — the numerical difference is the same)
$K_b$ is a property of the solvent alone, not the solute. For water, $K_b = 0.512\ \text{K kg mol}^{-1}$
$m$ is the molality of the solution

So if you dissolve 1 mole of a non-volatile solute in 1 kg of water, the boiling point rises by $0.512^\circ\text{C}$ — from $100^\circ\text{C}$ to $100.512^\circ\text{C}$ .

Watch out

Molality ( $m$ ) is not the same as molarity ( $M$ ). Molality uses mass of solvent (kg), molarity uses volume of solution (L). For dilute aqueous solutions they are numerically close, but in exact problems the distinction matters.

Why proportional to molality, not molarity?

Because boiling point elevation depends on the number of solute particles relative to the mass of solvent, not the volume of the solution. Temperature changes affect volume (and therefore molarity), but mass stays constant. Molality is temperature-independent, making it the natural choice for a property that itself depends on temperature.

A concrete example

Suppose you dissolve 58.44 g of NaCl (table salt, molar mass 58.44 g/mol) in 500 g of water. That is 1 mole of NaCl in 0.5 kg of water, so molality $m = 2\ \text{mol/kg}$ .

But NaCl dissociates in water into Na⁺ and Cl⁻ ions — two particles per formula unit. For ionic solutes, the effective number of particles is given by the van't Hoff factor $i$ . For NaCl, $i \approx 2$ . …

Part (b)Concept understanding — Raoult's Law

Raoult's Law: From Intuition to Precision

Imagine you have a beaker of pure water at room temperature. Some water molecules at the surface have enough energy to escape into the air above — that's vapour pressure. Now dissolve some sugar in that water. The sugar molecules take up space at the surface, blocking some water molecules from escaping. Fewer water molecules can leave the liquid per second, so the vapour pressure drops.

That's the core intuition: a non-volatile solute lowers the solvent's vapour pressure simply by getting in the way.

But what if both components can evaporate — say, a mixture of benzene and toluene? Then both kinds of molecules crowd the surface, and both contribute to the total vapour pressure. The question becomes: how much does each contribute?

The Precise Statement

For a solution of volatile liquids, Raoult's Law says:

p_i = x_i \, p_i^*

where:

$p_i$ = partial vapour pressure of component $i$ above the solution
$x_i$ = mole fraction of component $i$ in the liquid solution
$p_i^*$ = vapour pressure of pure component $i$ at the same temperature

The law applies to each volatile component separately. The total vapour pressure above the solution is simply the sum:

P_{\text{total}} = p_1 + p_2 = x_1 p_1^* + x_2 p_2^*

What This Means Physically

The mole fraction $x_i$ tells you the fraction of molecules at the surface that are of type $i$ . If half the molecules in the liquid are benzene ( $x_{\text{benzene}} = 0.5$ ), then roughly half the surface sites are occupied by benzene molecules. So the rate at which benzene escapes should be about half the rate from pure benzene — hence $p_{\text{benzene}} = 0.5 \times p_{\text{benzene}}^*$ .

This is a linear relationship: plot $p_i$ against $x_i$ , and you get a straight line from the origin (when $x_i = 0$ , $p_i = 0$ ) up to $p_i^*$ (when $x_i = 1$ , pure component).

Tip

Raoult's Law is an idealisation. It works best when the two liquids are chemically similar — same type of intermolecular forces (e.g., both non-polar, or both with similar hydrogen bonding). Benzene–toluene is a classic example. When the molecules interact very differently (like ethanol and water), the law fails — that's when you get deviations from Raoult's Law.

A Concrete Example

Suppose you mix 2 moles of benzene ( $p^* = 100$ mm Hg) with 3 moles of toluene ( $p^* = 40$ mm Hg) at 25°C.

Mole fractions:

$x_{\text{benzene}} = \frac{2}{2+3} = 0.4$
$x_{\text{toluene}} = \frac{3}{5} = 0.6$

Partial pressures:

$p_{\text{benzene}} = 0.4 \times 100 = 40$ mm Hg
$p_{\text{toluene}} = 0.6 \times 40 = 24$ mm Hg

Total vapour pressure: $40 + 24 = 64$ mm Hg

Notice: the total pressure is not a simple average of the pure pressures. It's a weighted average, with mole fractions as weights.

Why This Matters …

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Part (a): $\Delta T_b = i K_b m = 3\times0.52\times0.104 \approx 0.16\ \text{K}$ .

Part (b): By Raoult's law $P = 0.5(120)+0.5(160) = 140\ \text{mm Hg}$ .

Part (a)

Boiling-point elevation is a colligative property that depends on the number of solute particles. For an electrolyte we include the van't Hoff factor $i$ :

\Delta T_b = i\,K_b\,m

Van't Hoff factor: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$ produces $1+2 = 3$ ions, so with complete dissociation $i = 3$ .

Molality:

\text{moles } CaCl_2 = \frac{3}{111}=0.0270\ \text{mol}, \quad \text{water}=0.260\ \text{kg}

m = \frac{0.0270}{0.260}=0.1040\ \text{mol kg}^{-1}

Elevation: …

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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.

CBSE 2026Set A1 markMCQ
$Q. The elevation in boiling point produced by one molal solution of a solute in a solvent is called$
(a) Ebullioscopic constant
(b) Vapour pressure constant
(c) Cryoscopic constant
(d) None of these
›Reveal solutionSolution

The boiling-point elevation for a 1 molal solution equals Kb, the molal elevation (ebullioscopic) constant.

The elevation in boiling point is ΔTb = Kb x m. When molality m = 1, ΔTb = Kb, called the ebullioscopic constant (molal elevation constant). The cry …

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CBSE 2026Set ANNUAL1 mark
$Q. The unit of molal elevation constant is ______ (fill in the blank).$
›Reveal solutionSolution

The molal elevation constant Kb is defined by ΔTb = Kb·m, so its unit follows from rearranging: Kb = ΔTb/m.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following aqueous solutions will have the maximum boiling point?$
(a) 1% glucose
(b) 1% sucrose
(c) 1% NaCl
(d) 1% CaCl2
›Reveal solutionSolution
Boiling point elevation, delta-Tb = i x Kb x m, depends on the TOTAL number of solute particles, not the solute's identity - so for equal 1% mass solutions we must compare (mass/molar mass) x i for each.

For a fixed mass percentage (1% w/V), the effective molal concentration driving boiling-point elevation is proportional to (1/M) x i, where M is the molar mass and i is the van't Hoff factor (number of particles the solute dissociates into).

1% glucose (M = 180 g/mol, i = 1): (1/180) x 1 = 0.00556

1% sucrose (M = 342 g/mol, i = 1): (1/342) x 1 = 0.00292 …
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CBSE 2026Set SEM31 markMCQ
$Q. If the elevation in boiling point of a solution of 10 g of non-volatile, non-dissociative, non-associative solute in 100 g of water is dTb and if molar mass of the solute is 100 then Ebullioscopic constant of water 'Kb' is$
(a) 100 (dTb)
(b) dTb
(c) dTb / 100
(d) 10 (dTb)
›Reveal solutionSolution

Elevation in boiling point dTb = Kb x m. Here molality m = 1 mol/kg, so Kb = dTb. Correct option (b).

For a dilute solution of a non-volatile, non-dissociating solute: dTb = Kb x m, where m is molality.

Step 1 - moles of solute: n = mass / molar mass = 10 / 100 = 0.1 mol.

Step 2 - mass of solvent: 100 g water = 0.1 kg.

Step 3 - molality: m = n / (kg solvent) = 0.1 / 0.1 = 1 mol/kg. …

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CBSE 2025Set D1 markMCQ
$Q. Which among the following aqueous solutions has the highest boiling point?$
(a) 1% glucose
(b) 1% sucrose
(c) 1% NaCl
(d) 1% CaCl2
›Reveal solutionSolution
Boiling-point elevation depends on total particle concentration; 1% NaCl gives the most particles, so it has the highest boiling point.

Elevation of boiling point is a colligative property: delta Tb = i x Kb x molality, so the solution with the largest (i x moles of particles) for the same mass wins. For 1% w/w (same mass of solute per unit solvent), compute particles per gram = (i / molar mass):

1% glucose: non-electrolyte i = 1, M = 180 -> 1/180 = 0.0056

1% sucrose: non-electrolyte i = 1, M = 342 -> 1/342 = 0.0029

1% NaCl: i = 2, M = 58.5 -> 2/58.5 = 0.0342 …
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CBSE 2025Set ANNUAL1 mark
$Q. Write the mathematical form of Raoult's law.$
›Reveal solutionSolution

Raoult's law: for a solution of volatile liquids, each component's partial vapour pressure is proportional to its mole fraction, with the pure-component vapour pressure as the proportionality constant.

For a binary solution of two volatile liquids, components 1 and 2:

p1 = x1 . p1(deg)

p2 = x2 . p2(deg)

where p1(deg), p2(deg) are the vapour pressures of the pure components, and x1, x2 are their mole fractions in solution.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following aqueous solutions should have the highest boiling point?$
(a) 1.0 M NaOH
(b) 1.0 M Na2SO4
(c) 1.0 M NH4NO3
(d) 1.0 M KNO3
›Reveal solutionSolution
Boiling point elevation is a colligative property that depends on the total number of solute particles in solution, not their identity - so at equal molar concentration, the electrolyte that dissociates into the most ions gives the highest boiling point.

Elevation in boiling point: $\Delta T_b = i K_b m$ , where $i$ is the van't Hoff factor (number of particles produced per formula unit on dissociation).

At the same concentration (1.0 M) for all four solutes, compare $i$ :

NaOH -> Na+ + OH- : i = 2

Na2SO4 -> 2Na+ + SO4^2- : i = 3

NH4NO3 -> NH4+ + NO3- : i = 2 …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. An example of colligative property of solution is$
(a) density
(b) mass
(c) elevation of boiling point
(d) temperature
›Reveal solutionSolution

Colligative properties depend only on the number of solute particles present, not on their chemical nature.

Density, mass and temperature are ordinary physical quantities that do not depend on the number of dissolved particles - they are not colligative properties. Elevation of boiling point, however, depends directly on the mole fraction/molality of the solute particles (via $\Delta T_b = i K_b m$ ), making …

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CBSE 2025Set ANNUAL1 mark
$Q. Assertion [A] : On addition of non-volatile solute in a pure solvent, boiling point of the solution increases. Reason [R] : On addition of non-volatile solute in a pure solvent, vapour pressure of solution increases.$
›Reveal solutionSolution

Boiling-point elevation is real (Assertion true), but it happens because vapour pressure DECREASES (not increases) when a non-volatile solute is added — so the stated Reason is false.

By Raoult's law, adding a non-volatile solute lowers the mole fraction of solvent at the surface, which LOWERS the vapour pressure of the solution below that of the pure solvent. Because the solution's vapour pressure is now lower, it must be heated to a HIGHER temperature before its vapour pressure equals atmospheric pressure (i.e., before it boils) — this is exactly why boiling point rises ( …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The units of ebulloscopic constant is:$
(a) K kg mol⁻¹
(b) mol kg K⁻¹
(c) kg mol⁻¹ K⁻¹
(d) K mol kg⁻¹
›Reveal solutionSolution

The ebullioscopic constant (molal elevation constant) $K_b$ has units K kg mol⁻¹.

Boiling point elevation is given by:

$\Delta T_b = K_b \times m$

where $\Delta T_b$ is in kelvin and $m$ (molality) is in mol kg⁻¹. Rearranging, …

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$Q. Define Raoult's law for a solution containing a non-volatile solute.$
›Reveal solutionSolution

Raoult's law states that the vapour pressure of the solvent above a solution is directly proportional to the mole fraction of the solvent present.

Statement of Raoult's law

For a solution containing a non-volatile solute, Raoult's law states: the partial vapour pressure of the solvent ( $p_1$ ) over the solution is directly proportional to its mole fraction ( $x_1$ ) in the solution.

$p_1 \propto x_1 \quad\Rightarrow\quad p_1 = p_1^{\circ}\,x_1$

where $p_1^{\circ}$ is the vapour pressure of the pure solvent at that temperature. Since the solute is non-volatile, it contributes nothing to the vapour pressure, so $p_1$ IS the total vapour pressure of the solution.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Out of two 0.1 molal solution of glucose and of potassium chloride, why do 0.1 molal solution of potasium chloride have a higher boiling point ?$
›Reveal solutionSolution
Since boiling-point elevation depends on the number of dissolved particles, and KCl dissociates into two ions per formula unit while glucose stays as one particle, KCl produces roughly double the effect at the same molal concentration.

Boiling point elevation is a colligative property, given by $\Delta T_b = i K_b m$ , where $i$ is the van't Hoff factor (the effective number of particles each formula unit produces in solution).

Glucose is a non-electrolyte — it dissolves as intact molecules, so $i \approx 1$ .

Potassium chloride (KCl) is a strong electrolyte and dissociates essentially completely in water: $KCl \rightarrow K^+ + Cl^-$ , giving $i \approx 2$ . …
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Q.(a) Calculate the elevation of boiling point of a solution when 3 g of CaCl2CaCl_2CaCl2​ (Molar mass = 111 g mol−1^{-1}−1) was dissolved in 260 g of water, assuming that CaCl2CaCl_2CaCl2​ undergoes complete dissociation. (KbK_bKb​ for water = 0·52 K kg mol−1^{-1}−1)

Part (a)

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Q.(a) Calculate the elevation of boiling point of a solution when 3 g of $CaCl_2$ (Molar mass = 111 g mol $^{-1}$ ) was dissolved in 260 g of water, assuming that $CaCl_2$ undergoes complete dissociation. ( $K_b$ for water = 0·52 K kg mol $^{-1}$ )

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