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Q1.35:
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.

Solution :

Step 1: Theoretical Foundation of Henry's Law

Henry's Law states that the partial pressure of a gas in a vapor phase is proportional to the mole fraction of the gas in the solution. Mathematically, it is expressed as:

$P = K_H \cdot \chi$

Where:

  • $P$ = Partial pressure of the gas above the solution.
  • $K_H$ = Henry’s law constant.
  • $\chi$ = Mole fraction of the gas in the solution.

[Per Henry's Law for gas solubility in liquids]

Step 2: Identification of Given Variables

Based on the problem statement at $298\text{ K}$:

  • Partial Pressure ($P$) = $760\text{ mm Hg}$
  • Henry's Law Constant ($K_H$) = $4.27 \times 10^5\text{ mm Hg}$

Step 3: Rearrangement of Formula to Solve for Solubility ($\chi$)

To find the solubility of methane expressed as mole fraction ($\chi$), rearrange the equation:

$\chi = \frac{P}{K_H}$

Step 4: Calculation and Substitution

Substitute the given numerical values into the rearranged equation:

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

Perform the division:

$\chi = \frac{760}{427000}$

$\chi \approx 1.7798 \times 10^{-3}$

Visual Representation of Pressure-Solubility Relationship

Mole Fraction ($\chi$) Pressure ($P$) Slope = $K_H$

Step 5: Analysis of Results

The calculated mole fraction represents the solubility of methane in benzene. Since the mole fraction is a dimensionless quantity, it signifies the ratio of the number of moles of methane to the total number of moles in the solution. [By definition of mole fraction]

Final Solution: The solubility of methane in benzene at 298 K is $1.78 \times 10^{-3}$.


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