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31 August, 07:02

17. What mass of ethylene glycol, when mixed with 225 g H2O, will reduce the equilibrium vapor pressure of H2O from 1.00 atm to 0.800 atm at 100 °C? The molar masses of water and ethylene glycol are 18.02 g/mol and 62.07 g/mol, respectively. Assume ideal behavior for the solution. a. 15.6 g

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  1. 31 August, 08:36
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    The mass of ethyleneglycol that is been required to reduce the vapor pressure of H2O from 1.00 atm to 0.800 atm at 100 °C, is 193,6 g

    Explanation:

    ΔP = P° - X st

    where ΔP is P° - P' (Pressure pure sv - Pressure in sl)

    and X st is molar fraction from solute

    We replace the values:

    1,00 atm - 0,800 atm = 1 atm. Xst

    0,2 atm = 1 atm. Xst

    0,2 = X st (remember, molar fraction has no units)

    X st which is molar fraction from solute means: moles from solute / moles from solute + moles from solvent so

    0,2 = moles st / moles st + moles sv

    You have mass from solvent so let's get the moles:

    225 g / 18,02 g/m = 12,48 moles

    0,2 = moles st / moles st + 12, 48 moles

    0,2 (moles st + 12,48 moles sv) = moles st

    0,2 moles st + 2,496 moles sv = moles st

    2,496 moles = moles st - 0,2 moles st

    2,496 moles = 0,8 moles st

    2,496 moles / 0,8 moles = 3,12 moles

    So now let's get the mass with the molar mass

    3,12 moles x 62,07 g/moles = 193,6 g
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