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27 August, 20:08

At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226. torr. Suppose a solution is prepared by mixing 127. g of acetic acid and 141. g of methanol CH3OH. Calculate the partial pressure of acetic acid vapor above this solution. Round your answer to 3 significant digits.

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  1. 27 August, 20:36
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    The partial pressure of acetic acid is 73.5 torr

    Explanation:

    Step 1: Data given

    Total pressure is 226 torr

    mass of acetic acid = 126 grams

    mass of methanol = 141 grams

    Step 2: Calculate moles of acetic acid

    moles acetic acid = mass acetic acid / molar mass acetic acid

    moles acetic acid = 127 grams / 60.05 g/mol

    moles acetic acid = 2.115 moles

    Step 3: Calculate moles of methanol

    moles methanol = 141 grams / 32.04 g/mol

    moles methanol = 4.40 moles

    Step 4: Calculate total moles

    Total moles = moles of acetic acid + moles methanol

    Total moles = 2.115 moles + 4.40 moles

    Total moles = 6.515 moles

    Step 5: Calculate mole fraction of acetic acid

    2.115 moles / 6.515 moles = 0.325

    Step 6: Calculate partial pressure of acetic acid

    P (acetic acid) = 0.325 * 226

    P (acetic acid) = 73.45 torr ≈73.5

    We can control this by calculating the partial pressure of methanol

    mole fraction of methanol = (6.515-2.115) / 6.515 = 0.675

    P (methanol) = 0.675 * 226 = 152.55

    226 - 152.55 = 73.45 torr

    The partial pressure of acetic acid is 73.5 torr
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