Question 9 of 50intermediate🔧 ApplyNumerical2 marks

The partial pressure of ethane over a solution containing 6.56×6.56 \times 10^{-3} g of ethane is 1 bar. If the solution contains 5.00×5.00 \times 10^{-2}$$ g of ethane, then what shall be the partial pressure of the gas?

Correct Answer

The partial pressure of ethane shall be 7.62 bar.

Exercise: EXERCISES | Q: 1.13 | (Chapter: 28)
For More Understanding

Explanation

This question is based on Henry's Law, which states that the partial pressure of a gas above a solution is directly proportional to its mole fraction (or concentration) in the solution at constant temperature. Since the temperature and solvent remain the same, Henry's law constant (K_H) remains constant. Therefore, the ratio of partial pressure to mass of dissolved gas remains constant. By applying this proportionality, we can calculate the new partial pressure when the mass of dissolved ethane changes.

Solution Steps

  1. Step 1: According to Henry's Law, at constant temperature, the partial pressure of a gas is directly proportional to its concentration (mole fraction) in the solution.

    p ∝ x (mole fraction)

  2. Step 2: Since the solvent and temperature are constant, Henry's law constant (K_H) remains the same. If the amount of solvent is fixed, the mole fraction is proportional to the mass of dissolved gas.

    Therefore: p₁/m₁ = p₂/m₂

  3. Step 3: Given:

    • p₁ = 1 bar
    • m₁ = 6.56×106.56 \times 10⁻³ g
    • m₂ = 5.00×105.00 \times 10⁻² g
    • p₂ = ?

  4. Step 4: Applying the relation:

    p₂ = p₁ × (m₂/m₁)

    p₂ = 1 bar × (5.00×105.00 \times 10⁻²)/(6.56×106.56 \times 10⁻³)

    p₂ = 1 × (0.05/0.00656)

    p₂ = 1×7.621 \times 7.62

    p₂ = 7.62 bar

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