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.

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.

Since $x = \dfrac{n_{CH_4}}{n_{CH_4}+n_{water}} \approx \dfrac{n_{CH_4}}{n_{water}}$ (as $n_{CH_4} \ll n_{water}$ ): $n_{CH_4} = x \times n_{water} = 2.5\times10^{-3}\times55.56 = 0.1389$ mol.

Rounding, $n_{CH_4} \approx 1.38\times10^{-1}$ mol.

Common Mistakes

Forgetting to convert $K_H$ from kbar to bar (400 bar, not 0.4 bar) before dividing.

Misplacing the decimal point (choosing $1.38\times10^{-2}$ or $1.38\times10^{-3}$ instead of the correctly-scaled $1.38\times10^{-1}$ ).

✓Final answer
$1.38 \times 10^{-1}$

ANSWER: D
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 $K_H$ .

Helium is the least soluble of the four given gases in water at 293 K, so it has the highest Henry's law constant.

Common Mistakes

Confusing $K_H$ with solubility directly (thinking "highest $K_H$ " means "most soluble" — it's the opposite).

Assuming $\mathrm{H_2}$ , being the smallest molecule, is least soluble — in practice, among these four, He has the higher $K_H$ (lower solubility) in the standard reference data.

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

ANSWER: C
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: $\dfrac{x_{N_2}}{x_{O_2}}=\dfrac{K_{H,O_2}}{K_{H,N_2}}$ .

Substitute values: $\dfrac{34.86}{76.48}=0.4557$ .

Rounding, $\dfrac{x_{N_2}}{x_{O_2}}\approx0.45$ .

Common Mistakes

Inverting the ratio (using $K_{H,N_2}/K_{H,O_2}$ instead of the correct $K_{H,O_2}/K_{H,N_2}$ ).

Forgetting that a larger $K_H$ corresponds to lower solubility (smaller mole fraction) at a given pressure.

✓Final answer
$The correct option is (D) — 0.45.$

ANSWER: D
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: $x = \dfrac{p}{K_H} = \dfrac{1520}{4.27\times10^5}$ .

Compute: $x \approx 3.559\times10^{-3} \approx 3.56\times10^{-3}$ .

Common Mistakes

Forgetting to convert atm to mm Hg (unit mismatch with $K_H$ ).

Confusing mole fraction $x$ with molality or molarity.

✓Final answer
$3.56\times10^{-3}$

ANSWER: C
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 $= 200\ \mathrm{g}$ ): $n_{H_2O} = \dfrac{200}{18} = 11.11\ \mathrm{mol}$ .

Since $x_{CO_2} = \dfrac{n_{CO_2}}{n_{CO_2}+n_{H_2O}} \approx \dfrac{n_{CO_2}}{n_{H_2O}}$ (as $x_{CO_2}$ is small), we get $n_{CO_2} \approx x_{CO_2}\times n_{H_2O} = 2\times10^{-3}\times11.11 = 0.02222\ \mathrm{mol}$ .

Molarity $= \dfrac{n_{CO_2}}{V(\mathrm{L})} = \dfrac{0.02222}{0.200} = 0.1111\ \mathrm{mol\,L^{-1}} = 1.11\times10^{-1}\ \mathrm{mol\,L^{-1}}$ .

Common Mistakes

Forgetting to convert the volume from mL to L before dividing.

Using the total volume of solution instead of the amount of solvent when converting mole fraction to moles of solute — this only works because the solution is dilute so $n_{water}\approx n_{total}$ .

✓Final answer
$1.11 \times 10^{-1}$

ANSWER: B
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.

Mass of $O_2 = 1.111\times10^{-3}\ \text{mol}\times 32\ \text{g/mol} = 0.03556$ g $= 35.56$ mg.

This mass is dissolved in 1 kg (1 L) of water, so the concentration is $35.56$ mg/kg $= 35.56$ ppm $\approx 35.50$ ppm.

Common Mistakes

Forgetting that $K_H$ has units of pressure, so $x = p/K_H$ (not $p\times K_H$ ).

Using molarity instead of mole fraction directly, without converting through moles of water.

Forgetting the factor of 1000 (g to kg, or g to mg) when converting to ppm.

✓Final answer
$The correct option is (B) — 35.50.$

ANSWER: B
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): $Ar(40.39)>CO_2(1.67)>CH_4(0.413)>HCHO(1.83\times10^{-5})$ .

Solubility runs in exactly the opposite order (smallest KH = most soluble): $HCHO$ (most soluble) $>CH_4>CO_2>Ar$ (least soluble).

So, in INCREASING order of solubility: $Ar<CO_2<CH_4<HCHO$ .

This matches option (D).

Common Mistakes

Forgetting the inverse relationship and ranking solubility in the SAME order as KH instead of the reverse.

✓Final answer
$Ar < CO_2 < CH_4 < HCHO$

ANSWER: D
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 $= 1 - x_2 = 1 - 0.04 = 0.96$ .

Common Mistakes

Answering with the gas's new mole fraction (0.04, option A) instead of the water's mole fraction that the question actually asks for.

Forgetting that mole fraction of gas simply doubles under Henry's law (constant $K_H$ ) — some students mistakenly assume doubling pressure doubles the water's mole fraction instead, or leave it unchanged at 0.98 (option B), forgetting the pressure change entirely.

✓Final answer
$The correct option is (C) — 0.96.$

ANSWER: C
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.

Since $x_{CO_2} \approx \dfrac{n_{CO_2}}{n_{CO_2}+n_{water}} \approx \dfrac{n_{CO_2}}{n_{water}}$ (dilute gas): $n_{CO_2} = x_{CO_2}\times n_{water} = 3.034\times10^{-3}\times27.78 = 0.0843$ mol.

Mass of $CO_2 = 0.0843 \times 44 = 3.71$ g $\approx 3.7$ g.

Common Mistakes

Forgetting to convert atm to Pa before dividing by $K_H$ (which is given in Pa).

Using the total moles (gas + water) instead of approximating with just water moles — the small- $x$ approximation is standard here and gives a clean match.

✓Final answer
$The correct option is (A) — 3.7 g.$

ANSWER: A

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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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