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: Henry's law relates the partial pressure of a gas to its concentration in solution.

Henry's law states that the partial pressure of a gas is directly proportional to its mole fraction (or molality) in solution:

where $p$ is the partial pressure, $K_H$ is Henry's law constant, and $m$ is the molality.

Given:

• $K_H = 4.27 \times 10^5$ mm Hg (for molality scale)
• $p = 760$ mm Hg

Solution:

Rearranging Henry's law for molality: …

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:

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

1. 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)

2. Write Henry's law in the appropriate form.

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

Rearranging to solve for molality $m$:

3. Substitute the numerical values.

4. Calculate the molality.

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

Method: Direct Application of Henry’s Law

Henry’s Law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Formula

Where:

• $p$ = partial pressure of the gas (in mm Hg)
• $K_H$ = Henry’s law constant (in mm Hg per molality unit)
• $m$ = molality of the gas in solution (mol/kg of solvent)

Steps

1. Identify the given data

   • $K_H = 4.27 \times 10^5$ mm Hg
   • $p = 760$ mm Hg
   • Temperature = 298 K (constant, so Henry’s law applies)

2. Rearrange Henry’s law to solve for molality ($m$)

3. Substitute the values

4. Calculate …

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:

So:

✗ 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:

✗ 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:

• 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. …