Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does no

NCERT Exemplar · Q5

Q. Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon ____________.

(i) Temperature

(ii) Nature of solute

(iii) Pressure

(iv) Nature of solvent

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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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The solubility of a solid in a liquid is largely unaffected by changes in pressure because solids and liquids are nearly incompressible. The correct option is (iii).

When we talk about the maximum amount of a solid solute that can dissolve in a specified amount of a liquid solvent, we are discussing its solubility - a dynamic equilibrium between dissolution and crystallisation.

Let's examine each option:

(i) Temperature - Dissolution of most solids is endothermic, so increasing temperature shifts equilibrium toward more dissolution (a few solids are exothermic and behave oppositely, but temperature still matters). Solubility does depend on temperature.
(ii) Nature of solute - 'Like dissolves like': polar solutes dissolve well in polar solvents, non-polar in non-polar. Solubility does depend on the nature of the solute. …

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Here is the breakdown of the common mistakes students make on this specific concept, along with how to avoid them.

The Core Concept: Solubility of Solids in Liquids

The question tests your understanding of solubility — specifically, the factors that affect how much of a solid can dissolve in a liquid.

The correct answer is (iii) Pressure.

Why? Pressure has a significant effect on the solubility of gases in liquids (Henry's Law), but it has a negligible effect on the solubility of solids and liquids in liquids. Solids and liquids are nearly incompressible, so changing pressure doesn't change the space available for solute particles to fit.

Common Mistake #1: Confusing "Solid in Liquid" with "Gas in Liquid"

The Mistake:

Students often pick Pressure as a factor that does affect solubility, because they remember the general rule: "Solubility increases with pressure." This is true, but only for gases.

How to Avoid It:

Create a mental checklist by state of matter. When you see "solubility," immediately ask: Is the solute a solid, liquid, or gas?
Memorise the two key rules side-by-side:
- Solids in liquids: Solubility depends on Temperature and Nature (of both solute and solvent). Pressure is NOT a factor.
- Gases in liquids: Solubility depends on Temperature (inversely) and Pressure (directly, via Henry's Law). The nature of the gas and solvent also matters.

Common Mistake #2: Forgetting that "Nature" is a Factor

The Mistake:

Some students might incorrectly eliminate options (ii) or (iv), thinking that "nature" is too vague or not a real factor. They might think only temperature matters.

How to Avoid It:

Understand "Like Dissolves Like." The nature of the solute and solvent is the most fundamental factor. Polar solutes (like sugar) dissolve in polar solvents (like water). Non-polar solutes (like wax) dissolve in non-polar solvents (like kerosene). This is a direct dependence on nature.
Remember the definition: Solubility is a specific property of a pair of substances. You cannot talk about the solubility of "salt" without specifying the solvent (e.g., "in water" vs. "in oil").

Common Mistake #3: Misinterpreting the Effect of Temperature

The Mistake:

A student might think temperature is not a factor because they recall that for some solids (like $\ce{Ca(OH)2}$ or $\ce{Ce2(SO4)3}$ ), solubility decreases with an increase in temperature. They then incorrectly conclude temperature doesn't affect it.

How to Avoid It: …

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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 16 on this concept.

KCET 2025Set D-41 markMCQ
$Q. Variation of solubility with temperature T for a gas in liquid is shown by the following graphs. The correct representation is: (A) (B) (C) (D)$
›Reveal solutionSolution

Henry's law makes gas solubility fall as temperature rises — a straight line sloping DOWNWARD, not up, flat, or a hump.

Why solubility decreases with temperature for a gas. Dissolving a gas in a liquid is generally exothermic. As temperature rises, the dissolved gas molecules gain kinetic energy and increasingly escape back into the gas phase, so less gas stays dissolved at equilibrium — the everyday example is a fizzy drink going flat faster when warm. …
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KCET 2025Set D-41 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution

Use Henry's law to get the mole fraction of dissolved $\mathrm{N_2}$ , then convert that mole fraction to moles using the $\approx 55.5$ mol of water in 1 litre.

Step 1 — Henry's law.

The solubility of a gas in a liquid at a given temperature is proportional to its partial pressure above the liquid:
$p = K_H \cdot x$
where $x$ is the mole fraction of the gas in solution and $K_H$ is the Henry's-law constant. Rearranged:
$x = \frac{p}{K_H}$
Step 2 — Substitute (watch the units on $K_H$ ).

The constant is given as $76.48\ \mathrm{K\,bar}$ , i.e. $76.48$ kilobar $= 76.48\times10^3\ \mathrm{bar} = 76{,}480\ \mathrm{bar}$ (this is the standard NCERT value for $\mathrm{N_2}$ at 293 K).
$x_{\mathrm{N_2}} = \frac{0.987\ \mathrm{bar}}{76{,}480\ \mathrm{bar}} = 1.29\times10^{-5}$
A very small number — nitrogen is only sparingly soluble in water, as expected.

Step 3 — Moles of water in 1 litre.
$n_{\mathrm{H_2O}} = \frac{1000\ \mathrm{g}}{18\ \mathrm{g\,mol^{-1}}} = 55.5\ \mathrm{mol}$
Step 4 — Convert mole fraction to moles of $\mathrm{N_2}$ .

By definition
$x_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{N_2}}}{n_{\mathrm{H_2O}}}$
The approximation is excellent because $n_{\mathrm{N_2}} \lll n_{\mathrm{H_2O}}$ (we are about to find $n_{\mathrm{N_2}} \sim 10^{-4}$ against $55.5$ ). Hence
$n_{\mathrm{N_2}} = x_{\mathrm{N_2}} \times n_{\mathrm{H_2O}} = (1.29\times10^{-5})(55.5)$ $n_{\mathrm{N_2}} = 7.16\times10^{-4}\ \mathrm{mol}$
Step 5 — Check the options. …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{N}_2$
›Reveal solutionSolution
Henry’s law says the mole fraction of a dissolved gas is proportional to its partial pressure times its Henry’s constant. Using the given atmospheric ratio and Henry’s constants, the dissolved N₂:O₂ ratio comes out to about 2:1, so option (B) is correct.

Concept & Intuition

Rainwater is in contact with air, so gases dissolve according to Henry’s law:
$x_{\text{gas}} = \frac{p_{\text{gas}}}{K_H}$
where $x_{\text{gas}}$ is the mole fraction in the liquid, $p_{\text{gas}}$ is the partial pressure in the gas phase, and $K_H$ is Henry’s constant (here given in kbar). The atmospheric ratio N₂:O₂ is 4:1 by volume, which means the partial pressures are in the same ratio (since total pressure ≈ 1 atm, but we only need the ratio). The dissolved ratio is not simply 4:1 because O₂ dissolves more readily (lower $K_H$ ). We must compute the ratio of $p/K_H$ for each gas.

Step-by-step

Set up partial pressures In dry air, N₂ and O₂ are in volume ratio 4:1. Since partial pressure is proportional to mole fraction in the gas,

$p_{\mathrm{N_2}} : p_{\mathrm{O_2}} = 4 : 1$
Let $p_{\mathrm{O_2}} = P$ , then $p_{\mathrm{N_2}} = 4P$ .

Apply Henry’s law for each gas Henry’s law:

$x_{\mathrm{N_2}} = \frac{p_{\mathrm{N_2}}}{K_{H,\mathrm{N_2}}}, \quad x_{\mathrm{O_2}} = \frac{p_{\mathrm{O_2}}}{K_{H,\mathrm{O_2}}}$
Given $K_{H,\mathrm{N_2}} = 76.48$ kbar and $K_{H,\mathrm{O_2}} = 34.86$ kbar.

Find the ratio of dissolved mole fractions

$\frac{x_{\mathrm{N_2}}}{x_{\mathrm{O_2}}} = \frac{p_{\mathrm{N_2}} / K_{H,\mathrm{N_2}}}{p_{\mathrm{O_2}} / K_{H,\mathrm{O_2}}} = \frac{p_{\mathrm{N_2}}}{p_{\mathrm{O_2}}} \cdot \frac{K_{H,\mathrm{O_2}}}{K_{H,\mathrm{N_2}}}$
Substitute the partial pressure ratio $4/1$ :
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COMEDK 2024Set 2024-E1 markMCQ
$_2$
›Reveal solutionSolution

Henry's constant equals the slope/intercept of the partial-pressure line; the steepest line (S) has the highest $K_H$ and the flattest (Q) the lowest, giving $S>P>R>Q$ .

Henry's law: $p = K_H \cdot x_{\text{gas}}$ . On a partial-pressure vs mole-fraction plot the line for a gas has slope $K_H$ ; the higher its pressure-axis position/steepness, the larger $K_H$ .

From the graph the lines, ranked by steepness / pressure-axis intercept (highest→lowest), are: …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{K}_{\mathrm{H}}$
›Reveal solutionSolution

Convert the solubility to a mole fraction of $\text{O}_2$ , then apply Henry's law $p=K_H\,x$ to get $p\approx1.569\,\text{bar}$ .

Moles in 1 L of solution ( $\approx$ 1 L water, density $1\,\text{g/mL}$ ):
$n_{\text{O}_2}=\frac{0.08}{32}=2.5\times10^{-3}\,\text{mol},\qquad n_{\text{water}}=\frac{1000}{18}=55.56\,\text{mol}.$
Mole fraction of $\text{O}_2$ (its own amount is negligible in the denominator):

$x=\frac{2.5\times10^{-3}}{55.56}=4.5\times10^{-5}.$ …
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KCET 2023Set D-21 markMCQ
$Q. A 30% solution of hydrogen peroxide is (A) '30 volume' hydrogen peroxide (B) '10 volume' hydrogen peroxide (C) '50 volume' hydrogen peroxide (D) '100 volume' hydrogen peroxide$
›Reveal solutionSolution

Convert the 30% strength into moles of $\mathrm{H_2O_2}$ per 100 mL, use $2\mathrm{H_2O_2}\rightarrow 2\mathrm{H_2O}+\mathrm{O_2}$ to get the oxygen volume at STP, and express it per mL of solution — that number is the "volume strength".

1. What "volume strength" means

A hydrogen-peroxide solution labelled ' $x$ volume' liberates $x$ mL of $\mathrm{O_2}$ at STP from 1 mL of the solution on complete decomposition. So the whole problem is: how much $\mathrm{O_2}$ does 1 mL of a 30% solution give?

2. Decomposition stoichiometry
$2\,\mathrm{H_2O_2} \;\longrightarrow\; 2\,\mathrm{H_2O} + \mathrm{O_2}$
2 mol $\mathrm{H_2O_2}$ (i.e. $2\times 34 = 68$ g) give 1 mol $\mathrm{O_2}$ = 22.4 L at STP.

3. Take 100 mL of the solution

30% strength $\Rightarrow$ 30 g of $\mathrm{H_2O_2}$ in 100 mL of solution.
$n(\mathrm{H_2O_2}) = \frac{30}{34} = 0.882\ \text{mol}$ $n(\mathrm{O_2}) = \frac{0.882}{2} = 0.441\ \text{mol}$ $V(\mathrm{O_2}) = 0.441 \times 22.4 = 9.88\ \text{L} = 9882\ \text{mL (at STP)}$
4. Per mL of solution …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{K}_{\mathrm{H}}$
›Reveal solutionSolution
Statement (A) is incorrect: a higher Henry's constant $K_H$ means LOWER solubility, not higher.

Henry's law: $p = K_H \cdot x$ , where $x$ is the mole fraction of dissolved gas. Rearranging, $x = p/K_H$ , so for a fixed partial pressure, a larger $K_H$ gives a smaller $x$ (lower solubility).

Evaluating:

(A) "Higher $K_H$ ⇒ higher solubility" — INCORRECT; it is the reverse (higher $K_H$ ⇒ lower solubility). This is the classic wrong statement being tested.

(B) $K_H$ depends on the nature of gas and solvent — correct.

(C) $K_H$ is temperature dependent (it increases with temperature) — correct. …
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COMEDK 2023Set 2023-M1 markMCQ
$Q. Which of the following is incorrect regarding Henry's law? (A) Gas reacts with solvent chemically. (B) Pressure and concentrations are not too high. (C) Temperature is not too low. (D) Gas does not change its molecular state in solution i.e., neither dissociates nor associates.$
›Reveal solutionSolution
Henry's law holds only when the dissolved gas physically dissolves without chemical reaction, at moderate pressure/concentration and not-too-low temperature, and without changing its molecular state (no dissociation/association). The statement that the gas reacts with solvent chemically violates the law, so it is the incorrect one.

Henry's law ( $p = K_H \cdot x$ ) applies under the following conditions:

Pressure and concentration are not too high — (B) is a valid condition.

Temperature is not too low — (C) is a valid condition.

The gas does not change its molecular state in solution, i.e. it neither dissociates nor associates — (D) is a valid condition. …
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KCET 2022Set B-31 markMCQ
$CO_2$
›Reveal solutionSolution

The textbook-intended property is $CO_2$ 's low solubility in water, not its acidic character on its own — that low solubility is exactly what makes the bicarbonate buffer system (blood pH regulation, ocean carbon cycling) possible.

Why not the other options. $CO_2$ IS colourless/odourless and does have some compressibility, but neither of those properties explains its biological/geochemical significance. "Acidic nature" (forming carbonic acid) is real, but it's a consequence of how $CO_2$ behaves in water, not the root property being tested here. …
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KCET 2022Set B-31 markMCQ
$Q. Solubility of a gas in a liquid increases with (A) increase of P and decrease of T (B) decrease of P and decrease of T (C) increase of P and increase of T (D) decrease of P and increase of T$
›Reveal solutionSolution

Henry's law makes gas solubility rise with pressure, and because dissolution of a gas is exothermic, Le Chatelier's principle makes it rise as temperature falls.

Step 1 — The pressure dependence: Henry's law.
$p = K_H \cdot x$
where $p$ is the partial pressure of the gas above the solution, $x$ its mole fraction in solution, and $K_H$ the Henry's-law constant. Rearranged:
$x = \frac{p}{K_H}$
So solubility $x$ is directly proportional to pressure. Physically: higher pressure means more gas molecules striking the liquid surface per second, so more of them get captured until a new equilibrium is reached.

$\Rightarrow$ Increase P $\Rightarrow$ increase solubility.

(Everyday proof: a soda bottle is sealed under high $CO_2$ pressure; the instant you open it and the pressure drops, dissolved $CO_2$ fizzes out.)

Step 2 — The temperature dependence: Le Chatelier.

Dissolution of a gas in a liquid is an exothermic process, because the gas molecules lose their kinetic freedom and are stabilised by solvent interactions:

$\text{Gas} + \text{Solvent} \rightleftharpoons \text{Solution} + \text{Heat} \qquad (\Delta H < 0)$ …
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COMEDK 2022Set 20221 markMCQ
$Q. Which of the following does not affect solubility of a gas in liquid? (A) Nature of gas and liquid (B) Pressure (C) Concentration (D) Temperature$
›Reveal solutionSolution
'Concentration' is not an independent variable here - the concentration of the dissolved gas IS the solubility (the quantity being measured), not a factor that determines it. So concentration does not affect the solubility of a gas in a liquid.

Concept: Solubility of a gas in a liquid.

The factors that govern how much gas dissolves in a liquid are:

Nature of the gas and of the solvent (like dissolves like; CO2 is far more soluble in water than O2 because it reacts/interacts with water).

Pressure of the gas above the liquid - Henry's law, p = K_H * x, so solubility rises with partial pressure. …
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KCET 2021Set B-21 markMCQ
$_2$
›Reveal solutionSolution

Get the partial pressure of N $_2$ from Dalton's law, convert it to a mole fraction with Henry's law, then convert that mole fraction into moles dissolved in 10 mol of water.

Step 1 — Partial pressure of N $_2$ (Dalton's law).

Henry's law uses the partial pressure of the gas, not the total pressure:
$p_{\mathrm{N_2}} = y_{\mathrm{N_2}} \times P_{\text{total}} = 0.8 \times 5 = 4\ \text{atm}.$
Step 2 — Henry's law.
$p_{\mathrm{N_2}} = K_H \, x_{\mathrm{N_2}} \quad\Longrightarrow\quad x_{\mathrm{N_2}} = \frac{p_{\mathrm{N_2}}}{K_H} = \frac{4}{1.0\times10^{5}} = 4\times10^{-5}.$
The large $K_H$ tells us N $_2$ is only sparingly soluble — the tiny mole fraction is expected.

Step 3 — Convert mole fraction to moles.

$x_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{N_2}}}{n_{\mathrm{H_2O}}}\quad (\text{since } n_{\mathrm{N_2}} \lll n_{\mathrm{H_2O}})$ …
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