Henry's law constant for solubility of methane in benzene is $4.2 \times 10^{-5}$ mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at (i) 750 mm Hg (ii) 840 mm Hg.

Henry's law constant for solubility of methane in benzene is 4.2 × 10⁻⁵ mm Hg at a particular constant tempera

Write Brief Answer · Q45

4.2 \times 10^{-5}

(i) 750 mm Hg

(ii) 840 mm Hg.

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

67% · 45/67 Questions

🔒 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 — 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 …

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

Step 1. Henry's law states $x_{solute}=\dfrac{p_{solute}}{K_H}$ , so for the same gas-solvent pair at a fixed temperature, solubility (mole fraction dissolved) is directly proportional to the partial pressure above the solution.

Step 2. Using the constant exactly as printed, $K_H=4.2\times10^{-5}$ mm Hg: at 750 mm Hg, $x_1=\dfrac{750}{4.2\times10^{-5}}$ ; at 840 mm Hg, $x_2=\dfrac{840}{4.2\times10^{-5}}$ .

Step 3. Both of these come out far greater than 1, which is not physically possible for a mole fraction (mole fraction must lie between 0 and 1) -- this signals that the printed constant's exponent is very likely a misprint in this edition (a Henry's law constant for a sparingly soluble gas like methane in a solvent is normally of order $10^{4}$ - $10^{6}$ mm Hg, i.e. a POSITIVE exponent, not $10^{-5}$ ).

Step 4. Whatever the true magnitude of $K_H$ , the RATIO between the two solubilities is unaffected by this, since it cancels out: $\dfrac{x_2}{x_1}=\dfrac{p_2/K_H}{p_1/K_H}=\dfrac{p_2}{p_1}=\dfrac{840}{750}=1.12$ . …

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

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.

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 …
Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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.

…

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

…

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

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

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

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

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/1/11 markMCQ
$Q. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to : (A) high atmospheric pressure (B) low temperature (C) low atmospheric pressure (D) both low temperature and high atmospheric pressure$
›Reveal solutionSolution
Henry’s Law states that the solubility of a gas in a liquid is directly proportional to its partial pressure. At high altitude, low atmospheric pressure reduces the partial pressure of oxygen, lowering its concentration in blood and tissues. The correct answer is (C).

Why Henry’s Law is the key

The question is about oxygen concentration in blood — not about how much oxygen is in the air, but how much dissolves into the bloodstream. That’s governed by Henry’s Law, which describes the solubility of a gas in a liquid.
$C = k_H \cdot P$
For oxygen in blood, $k_H$ is fixed at body temperature. So the concentration of dissolved oxygen depends only on the partial pressure of oxygen in the air you breathe.

Step-by-step reasoning

At sea level, atmospheric pressure is about 1 atm. Oxygen makes up ~21% of air, so its partial pressure is $P_{O_2} \approx 0.21 \times 1 = 0.21$ atm. Henry’s Law then gives a normal dissolved oxygen concentration in blood.

At high altitude (e.g., 3000–5000 m), total atmospheric pressure drops significantly — to about 0.7 atm at 3000 m, and even lower higher up. The fraction of oxygen in air stays 21%, but the partial pressure becomes $P_{O_2} \approx 0.21 \times 0.7 = 0.147$ atm.

Applying Henry’s Law: Since $C = k_H \cdot P_{O_2}$ , a lower $P_{O_2}$ directly gives a lower $C$ . That’s why people at high altitude have low oxygen concentration in blood and tissues — not because the air has less oxygen percentage, but because the lower total pressure reduces oxygen’s partial pressure.

Check the options:

(A) High atmospheric pressure — Wrong. High pressure would increase oxygen solubility, not decrease it. …
Start your 14-day free trial to see the full solution →
CBSE 2023Set 56/2/11 markMCQ
$K_H$
›Reveal solutionSolution
Henry’s constant $K_H$ measures gas solubility in a liquid. Since solubility decreases when temperature rises, $K_H$ must increase with temperature. The correct option is (C).

Why Henry’s constant behaves this way

Henry’s law states that at a 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 mole fraction $x$ , the gas exerts a higher partial pressure — which implies the gas is less soluble (it prefers to stay in the vapour phase). Conversely, a smaller $K_H$ means the gas dissolves more readily.

So $K_H$ is an inverse measure of solubility: high $K_H$ → low solubility; low $K_H$ → high solubility.

How temperature affects solubility — and therefore $K_H$

Think about dissolving a gas. When a gas dissolves in a liquid, the process is generally exothermic (heat is released). This is because gas molecules must be surrounded by solvent molecules, and the new intermolecular attractions release energy.

Le Chatelier’s principle. For an exothermic dissolution, heat is a product:

$\text{Gas} + \text{solvent} \rightleftharpoons \text{solution} + \text{heat}$
Raising the temperature adds heat, which shifts the equilibrium backward — toward the undissolved gas. So solubility decreases as temperature rises.

Connect to $K_H$ . Since $K_H$ is inversely related to solubility, a decrease in solubility means $K_H$ must increase. Every textbook example confirms this: for oxygen in water, $K_H$ roughly doubles when going from 0 °C to 40 °C. …
Start your 14-day free trial to see the full solution →

←Previous question · evaluationQ44 — How many moles of solute particles are present in one litre of 10⁻⁴ M …→Next question · evaluationQ46 — The observed depression in freezing point of water for a particular so…

Q.Henry's law constant for solubility of methane in benzene is 4.2×10−54.2 \times 10^{-5}4.2×10−5 mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at

🎓Unlock everything free for 14 days

Q.Henry's law constant for solubility of methane in benzene is $4.2 \times 10^{-5}$ mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at

Unlock everything free for 14 days