At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is _____. (i) less than

NCERT Exemplar · Q3

Q. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

(i) less than the rate of crystallisation

(ii) greater than the rate of crystallisation

(iii) equal to the rate of crystallisation

(iv) zero

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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
Scuba diving	At depth, high pressure forces more N₂ into blood; rising too fast causes decompression sickness ("the bends")
Fish breathing	Oxygen dissolves in water at the surface (where partial pressure is highest); deeper water has less dissolved O₂
Altitude sickness	At high altitude, lower atmospheric pressure means less O₂ dissolves in your blood

The Key Takeaway

Important

Henry's Law is a proportionality: double the pressure above the liquid → double the gas dissolved in the liquid (at constant temperature). It's why carbonated drinks are bottled under pressure, why deep-sea divers must ascend slowly, and why a warm drink loses its carbonation faster.

The law is simple, but its consequences are everywhere — from the soda in your hand to the air you breathe at different altitudes.

Henry's law is a key quantitative concept in the NCERT/CBSE Class 12 Chemistry chapter on Solutions, and ‘Henry's law formula’ or ‘Henry's law numericals’ are common important-question searches for board exams, JEE Main and NEET. Its real-world applications, like gas solubility in carbonated drinks and blood at altitude, make it a favourite for application-based competitive-exam questions.

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

This is analogous to Raoult's Law for ideal solutions, but for a solute gas rather than a solvent.

Key Exam Points

Henry's Law works best for dilute solutions (low $x$ )
$k_H$ increases with temperature — gases become less soluble as temperature rises
$k_H$ is different for each gas-liquid pair — e.g., $CO_2$ in water vs $O_2$ in water
The law fails if the gas reacts chemically with the solvent (e.g., $HCl$ in water)

Quick Example

If $k_H = 3.0 \times 10^4 \text{ atm}$ for $O_2$ in water at 25°C, and the partial pressure of $O_2$ in air is 0.21 atm:

x = \frac{P}{k_H} = \frac{0.21}{3.0 \times 10^4} = 7.0 \times 10^{-6}

This tiny mole fraction explains why fish need gills to extract enough oxygen from water!

Bottom line: Henry's Law is a direct consequence of dynamic equilibrium at the gas-liquid interface, where the rates of dissolution and escape balance each other linearly.

At equilibrium, opposing processes occur at equal rates; dissolution and crystallisation balance perfectly, giving (iii).

Understanding Dynamic Equilibrium

Equilibrium in chemistry is not a static, frozen state - it's a dynamic balance. When a solid dissolves in a liquid, two processes compete:

Dissolution: solid particles leave the crystal lattice and enter the solution
Crystallisation: dissolved particles return to the solid phase

Initially, only dissolution occurs. As concentration rises, crystallisation begins too.

Reaching Equilibrium

Early stage: Rate of dissolution > rate of crystallisation - net dissolution continues.
Equilibrium: Rate of dissolution = rate of crystallisation - the solution becomes saturated; concentration stays constant, but particles continuously exchange between phases.
The key insight: equilibrium does not mean nothing is happening - forward and reverse processes proceed at identical rates, so no net change occurs.

Watch out

A common mistake is thinking equilibrium means "everything stops." In reality both dissolution and crystallisation continue - they just cancel out macroscopically.

\text{Rate}_{\text{dissolution}} = \text{Rate}_{\text{crystallisation}}

✓Final answer

The correct option is (iii) : equal to the rate of crystallisation.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing “equilibrium” with “no change” → Choosing (iv) zero

Why it happens:

Students often think “at equilibrium, nothing happens.” They see the word equilibrium and assume the rate must be zero.

How to avoid:

Remember: Equilibrium is dynamic, not static.

At equilibrium, the net change is zero, but the forward and reverse processes continue at equal rates.
For dissolution: solid particles leave the surface (dissolve) and dissolved particles return to the surface (crystallise) at the same speed.
So the rate is not zero — it is equal to the rate of crystallisation.

Correct choice: (iii) equal to the rate of crystallisation.

Mistake 2: Thinking dissolution stops when solution is saturated

Why it happens:

Students believe that once a solution is saturated, no more solid can dissolve, so the dissolution rate becomes zero.

How to avoid:

Saturation means the concentration of dissolved solute is at its maximum at that temperature.
But molecules are still moving: some solid leaves the surface, some dissolved solute returns.
At saturation, the two rates are equal — dissolution continues, but crystallisation matches it exactly.

Key takeaway:

“Saturated” ≠ “dissolution stopped.” It means dissolution rate = crystallisation rate.

Mistake 3: Misreading “volatile liquid solvent” and overcomplicating

Why it happens:

The phrase “volatile liquid solvent” distracts students. They think volatility changes the equilibrium behaviour.

How to avoid:

Volatility of the solvent affects vapour pressure and boiling, but not the dissolution–crystallisation equilibrium of a solid solute.
The principle of dynamic equilibrium for dissolution is the same regardless of solvent volatility.
Ignore the “volatile” label — it’s a red herring. Focus on the solid–solution interface.

Mistake 4: Picking (i) or (ii) — thinking one rate is always higher

Why it happens:

Students confuse the direction of net change before equilibrium with the state at equilibrium.

How to avoid:

Before equilibrium (unsaturated solution): dissolution rate > crystallisation rate → net dissolving.
At equilibrium: rates are equal.
After equilibrium (supersaturated): crystallisation rate > dissolution rate → net crystallisation.

The question asks at equilibrium — so only (iii) is correct.

Quick Summary Table

Mistake	Wrong choice	Why it’s wrong	Correct reasoning
Equilibrium = no activity	(iv) zero	Equilibrium is dynamic	Rates are equal, not zero
Saturation = dissolution stops	(iv) zero	Saturation is dynamic	Dissolution continues at same rate as crystallisation
Distracted by “volatile”	Any	Volatility irrelevant here	Focus on solid–solution equilibrium
Confusing before/at equilibrium	(i) or (ii)	Those describe net change before equilibrium	At equilibrium, rates are equal

Final answer: (iii) equal to the rate of crystallisation.

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.

Ruling out the other shapes: an upward line (A) or hump (C) would mean solubility rises with heating, which is the opposite of the real gas-solubility trend; a flat line (B) would mean no temperature dependence at all, which contradicts the well-established Henry's-law relationship.

✓Final answer
$Option (D) — the downward-sloping line.$
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.

$7.16\times10^{-4}$ is option (C). Option (A), $0.716\times10^{-3}$ , is numerically the same number ( $0.716\times10^{-3} = 7.16\times10^{-4}$ ) — but the standard NCERT-printed form of this answer, and the form matching the option set's convention, is (C). Options (B) and (D) are out by a factor of ten either way, which is what you get by mishandling the $\mathrm{k\,bar}$ in $K_H$ or forgetting the 55.5.

(Physical note: the everyday consequence of this small solubility is decompression sickness — divers breathing high-pressure air dissolve extra $\mathrm{N_2}$ in their blood, and it bubbles out if they surface too fast.)

✓Final answer
$7.16 \times 10^{-4}$

ANSWER: C
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 $4/1$ :
$\frac{x_{\mathrm{N_2}}}{x_{\mathrm{O_2}}} = 4 \times \frac{34.86}{76.48}$

Calculate numerically

$\frac{34.86}{76.48} \approx 0.4558$
Then
$4 \times 0.4558 \approx 1.823$
So the ratio is roughly $1.823 : 1$ , which is very close to $2:1$ .

Select the closest option Among (A) 3:1, (B) 2:1, (C) 4:1, (D) 1:4, the value 1.823:1 is nearest to 2:1.

Watch out

A common mistake is to forget that Henry’s constants are different and simply take the atmospheric ratio 4:1 as the dissolved ratio. That would give option (C), which is incorrect because O₂ dissolves more easily.

Tip

Notice that the ratio of Henry’s constants is about 2.2 (76.48/34.86 ≈ 2.19), so the dissolved ratio is roughly 4 divided by 2.2 ≈ 1.8, confirming 2:1 is the best match.

✓Final answer
$The correct option is (B). ANSWER: B$
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:

$S$ — starts highest, descends most steeply → largest $K_H$ .

$P$ — next.

$R$ — lower.

$Q$ — starts lowest and is nearly flat → smallest $K_H$ .

Hence decreasing order of $K_H$ : $S > P > R > Q$ .

✓Final answer
$S > P > R > Q$
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):
$x=\frac{2.5\times10^{-3}}{55.56}=4.5\times10^{-5}.$
Henry's law with $K_H=34.86\,\text{kbar}=34860\,\text{bar}$ :
$p=K_H\,x=34860\times4.5\times10^{-5}=1.569\,\text{bar}.$
✓Final answer
$\text{O}_2$
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
$\text{Volume strength} = \frac{9882\ \text{mL of O}_2}{100\ \text{mL of solution}} \approx 98.8 \approx \mathbf{100}$
5. Shortcut worth remembering
$\text{Volume strength} \approx 11.2 \times \text{molarity}, \qquad M = \frac{30 \times 10}{34} = 8.82\ \mathrm{M}$ $\Rightarrow 11.2 \times 8.82 = 98.8 \approx 100$
Both routes agree: 30% $\mathrm{H_2O_2}$ is the familiar "100 volume" peroxide of the laboratory shelf. (The other classic pairing to remember: 3% $\mathrm{H_2O_2}$ = '10 volume', the antiseptic bottle.)

✓Final answer
$The correct option is (D) '100 volume' hydrogen peroxide .$

ANSWER: D
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.

(D) $K_H$ is the intended constant of proportionality; the wording targets the solubility relationship in (A) as the key error.

✓Final answer
$K_H$
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.

The gas must dissolve physically, not react chemically with the solvent (e.g. $\mathrm{NH_3}$ or $\mathrm{CO_2}$ reacting with water deviate from Henry's law).

Statement (A) says the gas reacts with solvent chemically, which is precisely the situation where Henry's law fails. Hence (A) is the incorrect statement.

✓Final answer
$The correct option is (A) — Gas reacts with solvent chemically.$
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.

The key property. $CO_2$ 's comparatively low solubility in water means it doesn't simply dissolve and disappear — it persists in equilibrium with dissolved $CO_2$ / $H_2CO_3$ / $HCO_3^-$ , which is exactly the buffering mechanism blood and ocean chemistry depend on, and the reservoir effect that matters for the global carbon cycle.

✓Final answer
$Option (A) — its low solubility in water.$
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:
$\text{Gas} + \text{Solvent} \rightleftharpoons \text{Solution} + \text{Heat} \qquad (\Delta H < 0)$
By Le Chatelier's principle, raising the temperature (adding heat) drives an exothermic equilibrium backwards — the gas escapes. Lowering the temperature drives it forward — more gas dissolves.

$\Rightarrow$ Decrease T $\Rightarrow$ increase solubility.

(Everyday proof: warm soda goes flat quickly; aquatic life suffers in warm water because dissolved $O_2$ is lower — the basis of thermal pollution.)

Step 3 — Combine.

Solubility of a gas increases with increase of P and decrease of T — which is exactly why gases are stored dissolved under high pressure at low temperature.

✓Final answer
$The correct option is (A) — increase of P and decrease of T.$

ANSWER: A
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.

Temperature - dissolution of a gas is exothermic, so by Le Chatelier's principle solubility falls as temperature rises (this is why boiled water tastes flat).

'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.

✓Final answer
$The correct option is (C) — Concentration$

ANSWER: C
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.
$x_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{N_2}}}{n_{\mathrm{H_2O}}}\quad (\text{since } n_{\mathrm{N_2}} \lll n_{\mathrm{H_2O}})$ $n_{\mathrm{N_2}} \approx x_{\mathrm{N_2}} \times n_{\mathrm{H_2O}} = 4\times10^{-5} \times 10 = 4.0\times10^{-4}\ \text{mol}.$
Step 4 — Check the approximation. $n_{\mathrm{N_2}} = 4\times10^{-4}$ vs. $10$ mol water — the denominator correction is one part in $25{,}000$ , entirely negligible, so $4.0\times10^{-4}$ stands.

✓Final answer
$4.0 \times 10^{-4}$

ANSWER: A

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Q.At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

Concept understanding — Henrys Law

Henry's Law: The Physics of "Fizz"

The Intuition: Gas Wants to Dissolve

The Precise Statement

What the Constant $k_H$ Tells You

Real-Life Examples

The Key Takeaway

Why this formula?

Henry's Law: Why the Formula Holds

Why This Linear Relationship Exists

1. Dynamic Equilibrium at the Interface

2. The Driving Force for Dissolution

3. The Driving Force for Escape

4. Equating the Two Rates

Why It's Linear (Not Exponential or Logarithmic)

Key Exam Points

Quick Example

Understanding Dynamic Equilibrium

Reaching Equilibrium

Concept: Dynamic Equilibrium in Solutions

Method: Dynamic Equilibrium Principle

Steps:

Common Mistakes & How to Avoid Them

Mistake 1: Confusing “equilibrium” with “no change” → Choosing (iv) zero

Mistake 2: Thinking dissolution stops when solution is saturated

Mistake 3: Misreading “volatile liquid solvent” and overcomplicating

Mistake 4: Picking (i) or (ii) — thinking one rate is always higher

Quick Summary Table

1. What "volume strength" means

2. Decomposition stoichiometry

3. Take 100 mL of the solution

4. Per mL of solution

5. Shortcut worth remembering

Q.At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

Understanding Dynamic Equilibrium

Reaching Equilibrium

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