$K_H$ value for Ar(g), $CO_2$(g), HCHO(g) and $CH_4$(g) are 40.39, 1.67, $1.83\times10^{-5}$ and 0.413 respectively. Arrange these gases in the order of their increasing solubility. (i) $HCHO < CH_4 < CO_2 < Ar$ (ii) $HCHO < CO_2 < CH_4 < Ar$ (iii) $Ar < CO_2 < CH_4 < HCHO$ (iv) $Ar < CH_4 < CO_2 < HCHO$

KH value for Ar(g), CO₂(g), HCHO(g) and CH₄(g) are 40.39, 1.67, 1.83×10⁻⁵ and 0.413 respectively. Arrange thes

NCERT Exemplar · Q26

K_H

(i)

HCHO < CH_4 < CO_2 < Ar

(ii)

HCHO < CO_2 < CH_4 < Ar

(iii)

Ar < CO_2 < CH_4 < HCHO

(iv)

Ar < CH_4 < CO_2 < HCHO

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

70% · 92/131 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 →

Henry's law constant $K_H$ is inversely proportional to solubility: a higher $K_H$ means lower solubility. Ranking the gases by increasing $K_H$ gives the order of increasing solubility: Ar < CO₂ < CH₄ < HCHO.

Understanding Henry's Law

Henry's law describes how gases dissolve in liquids at equilibrium. The relationship is captured by:

p = K_H \cdot x

where $p$ is the partial pressure of the gas above the solution, $x$ is its mole fraction in the liquid, and $K_H$ is Henry's law constant.

The key insight: for a fixed partial pressure, if $K_H$ is large, the mole fraction $x$ must be small to satisfy the equation. In other words, a high $K_H$ means the gas is reluctant to dissolve — it prefers the gas phase. Conversely, a low $K_H$ means the gas readily dissolves into the liquid.

Think of $K_H$ as a measure of how much "pressure" you need to force a given amount of gas into solution. The harder it is to dissolve (higher $K_H$ ), the less soluble the gas.

Step-by-Step Solution

List the given Henry's constants

Gas $K_H$ value
Ar 40.39
CO₂ 1.67
HCHO $1.83 \times 10^{-5}$
CH₄ 0.413
Recognize the inverse relationship

Solubility $\propto \frac{1}{K_H}$ . The gas with the smallest $K_H$ is the most soluble; the gas with the largest $K_H$ is the least soluble.
Rank the $K_H$ values in increasing order

Arranging from smallest to largest:

Gas	$K_H$ value
Ar	40.39
CO₂	1.67
HCHO	$1.83 \times 10^{-5}$
CH₄	0.413

1.83 \times 10^{-5} < 0.413 < 1.67 < 40.39

This corresponds to:

K_H(\text{HCHO}) < K_H(\text{CH}_4) < K_H(\text{CO}_2) < K_H(\text{Ar})

Invert to get increasing solubility …

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

🔍 The Core Concept (Why First)

Henry’s Law states:

P = K_H \cdot x

Where:

$P$ = partial pressure of gas above the liquid
$x$ = mole fraction of gas dissolved in the liquid
$K_H$ = Henry’s Law constant

Key relationship:

Higher $K_H$ → lower solubility (gas escapes more easily)
Lower $K_H$ → higher solubility (gas dissolves more easily)

So solubility is inversely proportional to $K_H$ .

✗ Mistake #1: Thinking higher $K_H$ means higher solubility

Why it happens:

Students see a large $K_H$ and assume “big number = more dissolved.” This is the most common error.

How to avoid:

Always remember:

$K_H$ is a measure of escaping tendency — not dissolving power.

Write this rule on your formula sheet:

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

✗ Mistake #2: Forgetting units or comparing $K_H$ values without checking scale

Why it happens:

Here, $K_H$ values are given in different orders of magnitude (40.39, 1.67, $1.83\times10^{-5}$ , 0.413). Students sometimes ignore the exponent and just compare the digits.

How to avoid:

Always write all $K_H$ values in scientific notation with the same exponent for easy comparison:

Ar: $40.39$ → $4.039 \times 10^1$
CO₂: $1.67$ → $1.67 \times 10^0$
CH₄: $0.413$ → $4.13 \times 10^{-1}$
HCHO: $1.83 \times 10^{-5}$

Now clearly:

$1.83 \times 10^{-5} < 4.13 \times 10^{-1} < 1.67 < 40.39$

✗ Mistake #3: Arranging in order of $K_H$ instead of solubility

Why it happens:

Students list gases from smallest $K_H$ to largest $K_H$ and think that’s the solubility order — but they forget to invert.

How to avoid:

Do it in two clear steps:

Arrange $K_H$ in increasing order:

HCHO ( $1.83\times10^{-5}$ ) < CH₄ ( $0.413$ ) < CO₂ ( $1.67$ ) < Ar ( $40.39$ )
Invert for solubility: …

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.

AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$K_H$
›Reveal solutionSolution
Applying Henry's law to find methane's mole fraction, then converting to moles dissolved in 1 L of water, gives about $1.38\times10^{-1}$ mol — option (D).

Concept and Intuition

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: $p = K_H x$ . A large $K_H$ (here 400 bar) means the gas is not very soluble, so only a small mole fraction dissolves under a given pressure. Once we know the mole fraction, and knowing that water vastly outnumbers the dissolved gas molecules, we can approximate moles of water as the total moles in solution to extract the actual moles of dissolved methane.

Step-by-Step Solution

Henry's law: $p = K_H x \Rightarrow x = \dfrac{p}{K_H} = \dfrac{1\ \text{bar}}{0.4\ \text{kbar}} = \dfrac{1}{400} = 2.5\times10^{-3}$ .

Moles of water in 1 L (mass ≈ 1000 g, taking density ≈ 1 g/mL): $n_{water} = \dfrac{1000}{18} = 55.56$ mol. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution
Henry's law constant increases as solubility decreases; helium, being the least soluble of the four gases in water, has the highest $K_H$ — option (C).

Concept and Intuition

Henry's law states $p = K_H \cdot x$ , where $p$ is the partial pressure of the gas and $x$ is its mole fraction dissolved in the liquid. A larger $K_H$ means a smaller mole fraction dissolves for the same partial pressure — i.e., higher $K_H$ = lower solubility. Noble gases like helium have very weak intermolecular (van der Waals) interactions with water and are notoriously poorly soluble, which is why divers use helium-based gas mixtures (to reduce the amount of gas — like nitrogen — that dissolves in blood and causes decompression sickness).

Step-by-Step Solution

Recall the qualitative solubility order in water for these gases: $\mathrm{O_2}$ is somewhat more soluble than $\mathrm{N_2}$ , and both are more soluble than the very inert, small, monoatomic He.

Since $K_H \propto \dfrac{1}{\text{solubility}}$ , the least soluble gas has the largest $K_H$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution
Henry's law with equal partial pressures gives a mole-fraction ratio inversely proportional to the Henry's law constants, $\approx0.45$ .

Concept and Intuition

Henry's law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction dissolved in the liquid: $p=K_H\,x$ . A larger $K_H$ means the gas is less soluble (needs a higher partial pressure to dissolve the same mole fraction), so for two gases at the same partial pressure, the one with the larger $K_H$ ends up with the smaller mole fraction dissolved.

Step-by-Step Solution

Write Henry's law for each gas: $p_{N_2}=K_{H,N_2}\,x_{N_2}$ and $p_{O_2}=K_{H,O_2}\,x_{O_2}$ .

Given $p_{N_2}=p_{O_2}$ (equal partial pressures), divide the two equations: $\dfrac{x_{N_2}}{x_{O_2}}=\dfrac{K_{H,O_2}}{K_{H,N_2}}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$4.27\times10^5$
›Reveal solutionSolution
This is a direct application of Henry's law relating partial pressure of a gas above a liquid to its mole fraction dissolved in it.

Concept and Intuition

Henry's law states $p = K_H\,x$ , where $p$ is the partial pressure of the gas, $x$ is its mole fraction in solution, and $K_H$ is the Henry's law constant (here expressed in mm Hg, so the pressure must also be converted to mm Hg for consistent units).

Step-by-Step Solution

Convert pressure to mm Hg: $2\text{ atm} = 2\times760 = 1520$ mm Hg.

Apply Henry's law: $x = \dfrac{p}{K_H} = \dfrac{1520}{4.27\times10^5}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$\mathrm{CO_2}$
›Reveal solutionSolution
Using Henry's law $p = K_H x$ to get the mole fraction of dissolved $\mathrm{CO_2}$ , then converting mole fraction to molarity via the moles of water in 200 mL, gives $\approx 1.11\times10^{-1}\ \mathrm{mol\,L^{-1}}$ .

Concept and Intuition

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: $p = K_H\,x_{gas}$ . This lets us find the mole fraction of dissolved gas directly from the applied pressure and the Henry's law constant. Since the mole fraction of a dilute solute is tiny, we can approximate the total moles of solution as just the moles of solvent (water), and convert mole fraction to concentration using the known amount of solvent present.

Step-by-Step Solution

Apply Henry's law: $x_{CO_2} = \dfrac{p}{K_H} = \dfrac{3.4}{1.7\times10^3} = 2\times10^{-3}$ .

Find moles of water in 200 mL of soda water (density $\approx 1\ \mathrm{g/mL}$ , so mass $= 200\ \mathrm{g}$ ): $n_{H_2O} = \dfrac{200}{18} = 11.11\ \mathrm{mol}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$K_H$
›Reveal solutionSolution
This is a direct Henry's law + unit-conversion problem: convert the given partial pressure into a mole fraction, then into ppm (mg solute per kg solvent). The answer is 35.50 ppm.

Concept and Intuition

Henry's law states that for a gas dissolved in a liquid at a given temperature, the partial pressure of the gas above the solution is proportional to the mole fraction of the gas dissolved in the liquid: $p = K_H\,x$ . A large $K_H$ (like 50 kbar here) means the gas is poorly soluble — it takes a huge partial pressure to force even a tiny mole fraction into solution, which is exactly the situation for $O_2$ in water. Once we know the mole fraction, we can convert it to ppm because for very dilute solutions the mole fraction is essentially the ratio of moles of solute to (moles of solute + moles of solvent) ≈ moles of solute / moles of solvent.

Step-by-Step Solution

Apply Henry's law: $x_{O_2} = \dfrac{p}{K_H} = \dfrac{1\ \text{bar}}{50000\ \text{bar}} = 2\times10^{-5}$ .

Since the solution is dilute, $x_{O_2} \approx \dfrac{n(O_2)}{n(H_2O)}$ .

In 1 L of water, mass $\approx 1000$ g, so $n(H_2O) = \dfrac{1000}{18} = 55.56$ mol.

$n(O_2) = x_{O_2}\times n(H_2O) = 2\times10^{-5}\times 55.56 = 1.111\times10^{-3}$ mol. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$K_H$
›Reveal solutionSolution
Since solubility varies inversely with Henry's law constant, ranking the given KH values from largest to smallest and reversing gives the increasing-solubility order Ar<CO2<CH4<HCHO. Answer (D).

Concept and Intuition

Henry's law states $p=K_H x$ , where $p$ is the gas's partial pressure above the solution and $x$ is its dissolved mole fraction. For a FIXED partial pressure, a larger $K_H$ forces a SMALLER equilibrium mole fraction $x=p/K_H$ — meaning the gas is less soluble. So solubility and $K_H$ are inversely related.

Step-by-Step Solution

List the given $K_H$ values: $Ar=40.39$ , $CO_2=1.67$ , $HCHO=1.83\times10^{-5}$ , $CH_4=0.413$ .

Sort from largest $K_H$ (least soluble) to smallest $K_H$ (most soluble): $Ar(40.39)>CO_2(1.67)>CH_4(0.413)>HCHO(1.83\times10^{-5})$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$Q.A gas X is dissolved in water at 2 bar pressure. Its mole fraction in the solution is 0.02. Find the mole fraction of water in the solution when the pressure of the gas is doubled at the same temperature. (A) 0.04 (B) 0.98 (C) 0.96 (D) 0.02$
›Reveal solutionSolution
This tests Henry's law ( $p = K_H x$ ): doubling the gas pressure doubles the gas's mole fraction in solution, so the water's mole fraction drops from $0.98$ to $0.96$ .

Concept and Intuition

Henry's law says the partial pressure of a gas above a dilute solution is directly proportional to its mole fraction in the solution: $p = K_H \cdot x_{\text{gas}}$ , with $K_H$ constant at a given temperature. So if pressure doubles, the mole fraction of dissolved gas also doubles (as long as $K_H$ doesn't change, i.e., temperature is constant, as stated). The mole fraction of water then adjusts to keep the two mole fractions summing to 1 (a binary solution of gas + water).

Step-by-Step Solution

Given: at $p_1 = 2$ bar, mole fraction of gas $x_1 = 0.02$ .

Apply Henry's law to find $K_H$ : $K_H = \dfrac{p_1}{x_1} = \dfrac{2}{0.02} = 100$ bar.

At the same temperature, $K_H$ is unchanged. New pressure $p_2 = 2 \times 2 = 4$ bar.

New mole fraction of gas: $x_2 = \dfrac{p_2}{K_H} = \dfrac{4}{100} = 0.04$ .

Since gas + water make up the whole solution, mole fraction of water $= 1 - x_2 = 1 - 0.04 = 0.96$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$CO_2$
›Reveal solutionSolution
Apply Henry's law to get the mole fraction of dissolved CO₂, then convert to mass using the moles of water present in 500 mL. Answer: 3.7 g.

Concept and Intuition

Henry's law states that the partial pressure of a dissolved gas is proportional to its mole fraction in solution: $p = K_H \cdot x$ . Once the (small) mole fraction of dissolved gas is known, it can be converted to moles using the (much larger) known amount of solvent, since $x \approx n_{gas}/n_{solvent}$ when the gas is dilute.

Step-by-Step Solution

Convert pressure: $P = 5\ \text{atm} = 5\times101325\ \text{Pa} = 506{,}625$ Pa.

Henry's law: $x_{CO_2} = \dfrac{P}{K_H} = \dfrac{506{,}625}{1.67\times10^8} = 3.034\times10^{-3}$ .

Moles of water in 500 mL (density $\approx 1$ g/mL): $n_{water} = \dfrac{500}{18} = 27.78$ mol. …
Start your 14-day free trial to see the full solution →

←Previous questionQ25 — On the basis of information given below mark the correct option. Infor…→Next questionQ27 — [Note: two or more options may be correct] Which of the following fact…

Q.KHK_HKH​ value for Ar(g), CO2CO_2CO2​(g), HCHO(g) and CH4CH_4CH4​(g) are 40.39, 1.67, 1.83×10−51.83\times10^{-5}1.83×10−5 and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

Understanding Henry's Law

Step-by-Step Solution

🎓Unlock everything free for 14 days

Q. $K_H$ value for Ar(g), $CO_2$ (g), HCHO(g) and $CH_4$ (g) are 40.39, 1.67, $1.83\times10^{-5}$ and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

Unlock everything free for 14 days