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A sparingly soluble gas (solute) is in equilibrium with a solvent at 10 bar. The mole fraction of the solvent in the gas phase is 0.01. At the operating temperature and pressure, the fugacity coefficient of the solute in the gas phase and the Henry’s law constant are 0.92 and 1000 bar, respectively. Assume that the liquid phase obeys Henry’s law.
The MOLE PERCENTAGE of the solute in the liquid phase, rounded to 2 decimal places, is _____. 
    Correct answer is '0.91'. Can you explain this answer?
    Verified Answer
    A sparingly soluble gas (solute) is in equilibrium with a solvent at 1...
    H = 1000 bar,  = 0.92 for dilute solution
    We can use Henry law for solute Hx = yP
    (1000 x = 0.92 1 − 0.01) (10)
    Here, y and x are mole fraction of solute in gas and liquid phase respectively
    x = 9.108 ×10−3
    Mole percentage of solute in liquid phase = 100 (9.108 × 10−3) = 0.91
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    A sparingly soluble gas (solute) is in equilibrium with a solvent at 1...
    Solution:

    Given data:
    - Equilibrium pressure: 10 bar
    - Mole fraction of solvent in gas phase: 0.01
    - Fugacity coefficient of solute in gas phase: 0.92
    - Henry's law constant: 1000 bar

    We are required to find the mole percentage of the solute in the liquid phase.

    Step 1: Calculate the fugacity of the solvent in the gas phase

    The fugacity (f) is related to the mole fraction (x) and partial pressure (P) by the equation:
    f = x * P

    Given that the mole fraction of the solvent in the gas phase is 0.01, the fugacity of the solvent can be calculated as:
    f_s = 0.01 * 10 bar = 0.1 bar

    Step 2: Calculate the fugacity of the solute in the gas phase

    The fugacity of the solute (f_sol) can be calculated using the fugacity coefficient (Φ_sol) and the fugacity of the solvent (f_s) as:
    f_sol = Φ_sol * f_s

    Given that the fugacity coefficient of the solute in the gas phase is 0.92, the fugacity of the solute can be calculated as:
    f_sol = 0.92 * 0.1 bar = 0.092 bar

    Step 3: Calculate the mole fraction of the solute in the liquid phase

    According to Henry's law, the mole fraction of the solute in the liquid phase (x_sol) is related to the fugacity of the solute in the gas phase (f_sol) and the Henry's law constant (H) as:
    f_sol = x_sol * H

    Rearranging the equation, we get:
    x_sol = f_sol / H

    Substituting the values, we have:
    x_sol = 0.092 bar / 1000 bar = 0.000092

    Step 4: Calculate the mole percentage of the solute in the liquid phase

    The mole percentage of the solute in the liquid phase can be calculated as:
    Mole percentage = x_sol * 100

    Substituting the value of x_sol, we get:
    Mole percentage = 0.000092 * 100 = 0.0092%

    Rounding off the mole percentage to 2 decimal places, we get:
    Mole percentage ≈ 0.01%

    Therefore, the correct answer is '0.01%'.
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    A sparingly soluble gas (solute) is in equilibrium with a solvent at 10 bar. The mole fraction of the solvent in the gas phase is 0.01. At the operating temperature and pressure, the fugacity coefficient of the solute in the gas phase and the Henry’s law constant are 0.92 and 1000 bar, respectively. Assume that the liquid phase obeys Henry’s law.The MOLE PERCENTAGE of the solute in the liquid phase, rounded to 2 decimal places, is_____.Correct answer is '0.91'. Can you explain this answer?
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