Question 44 of 50intermediate🔧 ApplyNumerical2 marks

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

Correct Answer

Given: Henry's law constant, K_H = 4.27×4.27 \times 10^{5}$$ mm Hg Pressure, p = 760 mm Hg

Formula: According to Henry's law: p = K_H × x where x = mole fraction of gas in solution (measure of solubility)

Calculation: x = p/K_H x = 760 mm Hg / (4.27×4.27 \times 10^{5} mm Hg) x = 760 / 427000 x = $1.78 \times $10^{-3}

Answer: The solubility of methane in benzene at 298 K under 760 mm Hg is 1.78×1.78 \times 10^{-3}$$ (mole fraction).

Exercise: EXERCISES | Q: 1.35 | (Page 29)
For More Understanding

Explanation

This question applies Henry's law directly. The textbook states that 'the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution' and is expressed as p = K_H × x. The mole fraction of gas in solution is used as a measure of its solubility. By rearranging the equation to x = p/K_H and substituting the given values, we obtain the solubility as mole fraction.

Solution Steps

  1. Step 1: Identify given values - K_H = 4.27×4.27 \times 10^{5}$$ mm Hg, p = 760 mm Hg

  2. Step 2: Apply Henry's law formula: p = K_H × x

  3. Step 3: Rearrange to find mole fraction: x = p/K_H

  4. Step 4: Substitute values: x = 760/(4.27×4.27 \times 10^{5}$$)

  5. Step 5: Calculate: x = 760/427000 = 1.78×1.78 \times 10^{-3}$$

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