The following question is a case-based question. Read the case carefully and answer the questions that follow.

Question

Q. The following question is a case-based question. Read the case carefully and answer the questions that follow. According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this condition is fulfilled only in exceptional cases for mixtures (optical isomers, isotopic mixtures of an element, hydrocarbon mixtures). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This view is further supported by the fact that Raoult's law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent. Answer the following questions :

(a) Give one example of miscible liquid pair which shows negative deviation from Raoult's law. What is the reason for such deviation ?

(b)

(i) State Raoult's law for a solution containing volatile components.

(OR)

(b)

(ii) Raoult's law is a special case of Henry's law. Comment.

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Part (a)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 …

Part (b)Concept understanding — Henrys Law

Henry's Law: The Physics of "Fizz"

Imagine you open a cold bottle of soda. You hear that familiar psshhht sound. Bubbles rush out. Now think: why were those bubbles inside the bottle in the first place? The liquid wasn't boiling. The answer is Henry's Law.

The Intuition: Gas Wants to Dissolve

Gases are just molecules flying around. When a gas touches a liquid, some of those molecules get "trapped" inside the liquid — they dissolve. But here's the key: the more you push on the gas, the more of it gets forced into the liquid.

Think of a crowded bus. If you push more people toward the door (higher pressure), more people get squeezed inside. If you let the pressure off (open the bottle), people rush out. That's exactly what happens with gas and liquid.

In the soda bottle, carbon dioxide gas is pumped in at high pressure. That pressure forces a huge amount of CO₂ to dissolve into the liquid. When you open the bottle, the pressure above the liquid drops to normal air pressure. Suddenly, the liquid can't hold all that CO₂ anymore — so it escapes as bubbles. That's the fizz.

The Precise Statement

Henry's Law says:

C = k_H \cdot P

Where:

$C$ = concentration of the dissolved gas in the liquid (usually mol/L or g/L)
$P$ = partial pressure of that gas above the liquid (usually atm or kPa)
$k_H$ = Henry's law constant — a number that depends on the specific gas, the liquid, and the temperature

In words: At a constant temperature, the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid.

What the Constant $k_H$ Tells You

$k_H$ is not universal. It's different for every gas-liquid pair. For example:

CO₂ in water has a certain $k_H$
O₂ in water has a different $k_H$ (smaller — oxygen doesn't dissolve as easily)

Temperature matters too. Higher temperature means lower $k_H$ — gases become less soluble in hot liquids. That's why a warm soda goes flat faster than a cold one.

Watch out

Henry's Law works only for dilute solutions and non-reacting gases. If the gas reacts chemically with the liquid (like HCl gas dissolving in water to form hydrochloric acid), Henry's Law does not apply — the concentration will be much higher than predicted.

Real-Life Examples

Situation	What Henry's Law explains
Soda fizz	High pressure forces CO₂ in; releasing pressure lets it out

Why this formula?

Henry's Law: Why the Formula Holds

Henry's Law describes the solubility of a gas in a liquid at a constant temperature. The key formula is:

P = k_H \cdot x

Where:

$P$ = partial pressure of the gas above the liquid
$x$ = mole fraction of the gas dissolved in the liquid
$k_H$ = Henry's constant (depends on gas, liquid, and temperature)

Why This Linear Relationship Exists

1. Dynamic Equilibrium at the Interface

Imagine a gas above a liquid. At the molecular level:

Gas molecules constantly strike the liquid surface and dissolve
Dissolved molecules constantly escape back into the gas phase

At equilibrium, the rate of dissolution equals the rate of escape. This is a dynamic balance, not a static one.

2. The Driving Force for Dissolution

The rate at which gas molecules enter the liquid depends on:

How many gas molecules hit the surface — this is proportional to the partial pressure $P$ of the gas
How easily they dissolve — this is captured by $k_H$

So:

\text{Rate}_{\text{dissolve}} \propto P

3. The Driving Force for Escape

The rate at which dissolved molecules leave the liquid depends on:

How many dissolved molecules are near the surface — this is proportional to the mole fraction $x$ of the gas in the liquid
How easily they escape — also captured by $k_H$

So:

\text{Rate}_{\text{escape}} \propto x

4. Equating the Two Rates

At equilibrium:

\text{Rate}_{\text{dissolve}} = \text{Rate}_{\text{escape}}

Therefore:

P \propto x

Introducing the proportionality constant $k_H$ :

P = k_H \cdot x

Why It's Linear (Not Exponential or Logarithmic)

The linearity arises because:

No saturation effects at low concentrations — the molecules don't "crowd" each other
Ideal behavior is assumed — gas molecules don't interact strongly with each other or with the solvent
Temperature is constant — $k_H$ doesn't change …

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Part (a): chloroform-acetone shows negative deviation (stronger A-B H-bonding); Raoult's law for volatile components is $p_A=x_A p_A^\circ$ , $p_B=x_B p_B^\circ$ . Part (b): Raoult's law is Henry's law with $K_H=p_i^\circ$ ; ideal solutions have $\Delta H_{mix}=0$ and $\Delta V_{mix}=0$ .

Part (a)

(a) Negative deviation from Raoult's law. A classic miscible pair is chloroform + acetone (another is nitric acid + water). Negative deviation means the observed vapour pressure is below the Raoult prediction. It arises when the A-B intermolecular attractions are stronger than the A-A and B-B attractions in the pure liquids: chloroform's C-H forms a hydrogen bond to acetone's carbonyl oxygen. The tighter binding lowers each component's tendency to vaporise, so $p_{obs}<p_{Raoult}$ ; such mixing is exothermic ( $\Delta H_{mix}<0$ ) with a small volume contraction ( $\Delta V_{mix}<0$ ).

(b)(i) Raoult's law for volatile components. For a binary solution of two volatile liquids A and B, the partial vapour pressure of each component is proportional to its mole fraction in the liquid, the constant being the pure-liquid vapour pressure:

$p_A = x_A p_A^\circ, \qquad p_B = x_B p_B^\circ$ …

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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 25 on this concept.

CBSE 2026Set V11 markMCQ
$Q. The percentage of helium filled in the tanks used by most scuba divers to dilute air in deep dives$
(a) 32.1
(b) 11.7
(c) 74.2
(d) 56.2
›Reveal solutionSolution

The air in a deep-diving scuba tank is diluted with about 11.7% helium.

At the high pressures experienced in deep dives, the solubility of atmospheric gases in blood increases (Henry's law, $p = K_H\,x$ ). Nitrogen in particular dissolves and, on rapid ascent, is released as bubbles causing the painful and dangerous condition known as bends. To minimise this, the air supplied to divers is diluted with the sparingly-soluble, chemically inert gas hel …
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CBSE 2026Set ANNUAL1 mark
$Q. On increasing temperature, solubility of gases in liquids ______ (fill in the blank).$
›Reveal solutionSolution

Dissolution of a gas in a liquid is an exothermic process, so by Le Chatelier's principle, raising the temperature shifts the equilibrium back towards the gas phase, lowering solubility.

…

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CBSE 2025Set 56/4/11 markMCQ
$K_H$
›Reveal solutionSolution
Henry’s constant $K_H$ is inversely related to gas solubility — a higher $K_H$ means lower solubility. So the correct option is (B).

Why Henry’s Law works this way

Henry’s Law describes the relationship between the partial pressure of a gas above a liquid and its concentration in the liquid. The law is written as:
$p = K_H \cdot x$
where $p$ is the partial pressure of the gas, $x$ is its mole fraction in the solution, and $K_H$ is Henry’s constant.

The key intuition: $K_H$ is essentially a resistance to dissolution. A gas that dissolves easily (high solubility) will need only a small partial pressure to achieve a given concentration — so $K_H$ is small. Conversely, a gas that barely dissolves (low solubility) needs a large partial pressure to force even a tiny amount into solution — so $K_H$ is large.
$p = K_H \cdot x \quad \Rightarrow \quad K_H = \frac{p}{x}$
Step-by-step reasoning

Interpret the equation For a fixed partial pressure $p$ , the mole fraction $x$ of the dissolved gas is $x = p / K_H$ . Since $p$ is constant, $x$ (which measures solubility) is inversely proportional to $K_H$ :

$x \propto \frac{1}{K_H}$

Relate $K_H$ to solubility

High solubility → large $x$ → small $K_H$

Low solubility → small $x$ → large $K_H$

Check the options

(A) says $K_H$ is greater for gases with higher solubility — this is the opposite of what we just found.

(B) says $K_H$ is greater for gases with lower solubility — this matches the inverse relationship. …
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CBSE 2025Set 56/5/11 markMCQ
$K_H$
›Reveal solutionSolution

Henry’s law constant $K_H$ actually increases with temperature, not decreases — so Assertion (A) is false. The Reason (R) is true: gas solubility does fall as temperature rises. The correct choice is (D).

Why this question trips students up

The trap here is subtle. Most of us remember that “solubility of gases decreases with temperature” — that’s drilled into us from everyday life (a cold soda fizzes more than a warm one). So Reason (R) feels solid.

But the Assertion talks about Henry’s law constant $K_H$ , not solubility directly. The two are inversely related, and that inverse relationship flips the temperature dependence. Let’s walk through it carefully.

Step-by-step reasoning

1. Recall Henry’s law

Henry’s law states that at constant temperature, the partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid:
$p = K_H \cdot x$
Here $K_H$ is the Henry’s law constant. A larger $K_H$ means that for the same partial pressure, the gas dissolves less (smaller $x$ ). So $K_H$ is a measure of resistance to dissolution — it’s the opposite of solubility.
$p = K_H \cdot x \quad \Rightarrow \quad x = \frac{p}{K_H}$
2. What happens to solubility when temperature rises?

Dissolving a gas in a liquid is generally exothermic (heat is released). Le Chatelier’s principle tells us that raising temperature shifts equilibrium to favour the endothermic direction — which is the gas coming out of solution. So solubility $x$ decreases.

That makes Reason (R) true.

3. Now connect $K_H$ to temperature …
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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 mark
$Q. State Henry's law.$
›Reveal solutionSolution

Henry's law states that at constant temperature, the solubility (mole fraction) of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.

Statement: The partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas (x) dissolved in the solution, at a given temperature:

$p = K_H \cdot x$

where $K_H$ is the Henry's law constant, which is specific to the gas-solvent pair and depends on temperature.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Dissolution of a gas in liquid is a process$
(a) with increase in enthalpy
(b) with no change in enthalpy
(c) with decrease in enthalpy
(d) for which enthalpy change cannot be predicted
›Reveal solutionSolution

Dissolving a gas in a liquid is an exothermic process, so it occurs with a decrease in enthalpy.

When a gas dissolves in a liquid, gas molecules that were freely moving and widely separated come close to the solvent molecules and get surrounded by them (solvation). New, relatively strong solute–solvent (gas–liquid) interactions form as the gas molecules are trapped between solvent molecules. Energy is released when these new attractive interactions form, exactly as heat is released when a gas condenses.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Define Raoult's law for a solution containing a non-volatile solute.$
›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 markMCQ
$Q. The Kₕ values of Ar(g), CO2(g), HCHO(g) and CH4(g) are 40.39, 1.67, 1.83 \times 10⁻⁵ and 0.413 respectively. The correct order of their solubility is...$
(a) HCHO < CH4 < CO2 < Ar
(b) HCHO < CO2 < CH4 < Ar
(c) Ar < CO2 < CH4 < HCHO
(d) Ar < CH4 < CO2 < HCHO
›Reveal solutionSolution

Henry's law constant $K_H$ and gas solubility are inversely related — a HIGHER $K_H$ means LOWER solubility.

By Henry's law, $p = K_H \cdot x$ , where $x$ is the mole fraction of dissolved gas. For a fixed partial pressure $p$ , a larger $K_H$ forces a smaller mole fraction $x$ to dissolve — so solubility falls as $K_H$ rises. Given: $K_H$ (Ar) = 40.39, $K_H$ (CO2) = 1.67, $K_H$ (CH4) = 0.413, $K_H$ (HCHO) = 1.83×10⁻⁵. Ranking $K_H$ from highest to lowest: A …

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CBSE 2024Set B1 mark
$Q. Fill in the blank: The solubility of gas in a liquid is determined by ______ law.$
›Reveal solutionSolution

Henry's law states that at constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas dissolved in the liquid.

Henry's law: $p = K_H \cdot x$ , where p is the partial pressure of the gas above the solution, x is the mole fraction of the dissolved gas in the liquid, and $K_H$ is the Henry's law constant (specific to the gas-solvent pair and temperature). A higher $K_H$ means lower solubility for a given pressure. T …

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CBSE 2024Set ANNUAL1 mark
$Q. The mathematical form of Henry's Law is ______.$
›Reveal solutionSolution

Henry's Law states that the partial pressure of a gas in the vapour phase is directly proportional to its mole fraction dissolved in the solution.

Mathematically, Henry's Law is written as: p = KH . x …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The value of Henry's constant K_H :$
(a) Increases with increase in temperature
(b) Decreases with increase in temperature
(c) Remains constant
(d) First increases, then decreases
›Reveal solutionSolution

Gas solubility falls as temperature rises, and since $K_H$ is inversely related to solubility, $K_H$ rises with temperature.

Henry's law states $p = K_H \cdot x$ , where $p$ is the partial pressure of the gas above the solution and $x$ is its mole fraction dissolved. A larger $K_H$ means a smaller $x$ dissolves at the same pressure, i.e. lower solubility.

…

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Part (a)

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