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12 December, 04:13

At a certain temperature the vapor pressure of pure water is measured to be 212. mmHg. Suppose a solution is prepared by mixing 121. g of water and 59.6 g of methanol (CH3OH) Calculate the partial pressure of water vapor above this solution. Be sure your answer has the correct number of significant digits Note for advanced students: you may assume the solution is ideal mmHg

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  1. 12 December, 04:57
    0
    The vapor pressure will turn into a gass

    MmHUGejs
  2. 12 December, 05:03
    0
    The partial pressure of water vapor 166 mmHg

    Explanation:

    Step 1: Data given

    Vapor Pressure = 212 mmHg = 212 / 760 = 0.278947 atm

    Mass of water = 121 grams

    Mass of methanol = 59.6 grams

    Molar mass of water = 18.02 g/mol

    Molar mass of methanol = 32.04 g/mol

    Step 3: Calculate moles of water

    Moles water = mass water / molar mass water

    Moles water = 121 grams / 18.02 g/mol

    Moles water = 6.71 moles

    Step 4: Calculate moles methanol

    Moles methanol = 59.5 grams / 32.04 g/mol

    Moles methanol = 1.86 moles

    Step 5: Calculate total moles

    Total moles = moles water + moles methanol

    Total moles = 6.71 moles + 1.86 moles = 8.57 moles

    Step 6: Calculate mol fraction of water

    Mol fraction of water = 6.71 moles / 8.57 moles

    Mol fraction of water = 0.782

    Step 7: Calculate partial pressure of water vapor

    Partial pressure H2O = mol fraction water * vapor pressure

    Partial pressure H2O = 0.782 * 212 mmHg = 165.8 mmHg

    The partial pressure of water vapor is 165.8 ≈ 166 mmHg
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