Chemistry

Please help!
The vapor pressure of water at 35 oC is 42.175 mm Hg. The vapor pressure of ethyl alcohol (C2H5OH) at 35 C is 100.5 mm Hg. What is the vapor pressure of a solution prepared by dissolving 250 g of C2H5OH in 375 g of H2O?

The answer is supposed to be 48.3 mm Hg but I don't know how to get there!

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  1. mols EtOH = g/molar mass = ?
    mols H2O = g/molar mass = ?
    Total mols = mols EtOH + mols H2O = ?
    XEtOH = mols EtOH/total mols = ?
    XH2O = mols H2O/total mols = ?
    pEtOH = XEtOH*PoEtOH = ?
    pH2O = XH2O*PoH2O = ?
    Then Ptotal above the solution is pEtOH + pH2O
    Post your work if you get stuck.

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    DrBob222
  2. Here's my resulting answer but it isn't 48.3 mm Hg

    mols EtOH = g/molar mass = 5.43 mols
    mols H2O = g/molar mass = 20.81 mols
    Total mols = mols EtOH + mols H2O = 26.24 mols
    XEtOH = mols EtOH/total mols = 0.207 mol
    XH2O = mols H2O/total mols = 0.793 mol
    pEtOH = XEtOH*PoEtOH = 20.80 mmHg
    pH2O = XH2O*PoH2O = 33.45 mmHg

    Total Pressure = 54.25 mm Hg

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