Henry's law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Henry's law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubi

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4.27 \times 10^5

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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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Henry's law relates gas solubility to partial pressure: $p = K_H \cdot m$ . With $K_H = 4.27 \times 10^5$ mm Hg and $p = 760$ mm Hg, the molality of dissolved methane is $1.78 \times 10^{-3}$ mol/kg.

Henry's law tells us that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid. The proportionality constant $K_H$ depends on the gas-solvent pair and temperature. When $K_H$ is expressed as a "Henry's law constant for molality," the relationship takes the form:

p = K_H \cdot m

where $p$ is the partial pressure of the gas, $m$ is the molality of the dissolved gas, and $K_H$ has units of pressure per molality (here, mm Hg per mol/kg, which simplifies to mm Hg·kg/mol).

The physical intuition: a higher partial pressure "pushes" more gas molecules into solution. A larger $K_H$ means the gas is less soluble—you need more pressure to achieve the same molality.

Solution

Identify the given quantities.

We have:
- Henry's law constant: $K_H = 4.27 \times 10^5$ mm Hg (for molality)
- Partial pressure of methane: $p = 760$ mm Hg
- Temperature: $T = 298$ K (constant, so $K_H$ remains valid)
Write Henry's law in the appropriate form.

Since $K_H$ is given for molality, we use:

p = K_H \cdot m

Rearranging to solve for molality $m$ :

m = \frac{p}{K_H}

Substitute the numerical values.

m = \frac{760 \text{ mm Hg}}{4.27 \times 10^5 \text{ mm Hg}}

Calculate the molality.

$m = \frac{760}{4.27 \times 10^5} = \frac{760}{427000} \approx 1.78 \times 10^{-3} \text{ mol/kg}$ …

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Common Mistakes with Henry’s Law (Solubility Calculation)

The Problem in Brief

We are given:

Henry’s law constant ( $K_H$ ) = $4.27 \times 10^5$ mm Hg
Pressure ( $P$ ) = 760 mm Hg
Temperature = 298 K
Solvent = benzene
Solute = methane

We need solubility (molality, $m$ ).

Henry’s law:

P = K_H \cdot m

So:

m = \frac{P}{K_H} = \frac{760}{4.27 \times 10^5}

✗ Mistake #1: Forgetting the Units of $K_H$

What students do wrong:

They treat $K_H$ as dimensionless or assume it’s in atm, then convert incorrectly.

Why it happens:

$K_H$ is given in mm Hg, but many problems use atm. Students panic and convert unnecessarily.

How to avoid:

Always check the units of $K_H$ and $P$ — they must match.
Here, both are in mm Hg, so no conversion needed.
If they don’t match, convert one to match the other.

✓ Correct approach:

m = \frac{760 \text{ mm Hg}}{4.27 \times 10^5 \text{ mm Hg}} = 1.78 \times 10^{-3} \text{ mol/kg}

✗ Mistake #2: Confusing $K_H$ with Solubility

What students do wrong:

They think $K_H$ is the solubility, or they write $m = K_H \times P$ instead of $m = P / K_H$ .

Why it happens:

Henry’s law is often written as $P = K_H \cdot C$ , so students invert the relationship.

How to avoid:

Memorise the logic: Higher $K_H$ means lower solubility (gas escapes easily).
Write the formula clearly:

P = K_H \cdot m \quad \Rightarrow \quad m = \frac{P}{K_H}

Do a sanity check: $K_H$ is huge ( $4.27 \times 10^5$ ), so solubility should be tiny — and it is ( $\approx 0.0018$ molal).

✗ Mistake #3: Using the Wrong Form of Henry’s Law

What students do wrong:

They use $m = K_H \cdot P$ (inverse relationship) or apply $P = K_H \cdot x$ (mole fraction form) when molality is asked.

Why it happens:

Henry’s law has multiple forms — mole fraction, molarity, molality. Students mix them up.

How to avoid:

Read the question carefully: It says “molality” — so use the molality form. …

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Q.Henry's law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

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

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)

Solution

Method: Direct Application of Henry’s Law

Formula

Steps

Common Mistakes with Henry’s Law (Solubility Calculation)

The Problem in Brief

✗ Mistake #1: Forgetting the Units of $K_H$

✗ Mistake #2: Confusing $K_H$ with Solubility

✗ Mistake #3: Using the Wrong Form of Henry’s Law

Q.Henry's law constant for the molality of methane in benzene at 298 K is 4.27×1054.27 \times 10^54.27×105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution

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Q.Henry's law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

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