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Calculate the partial pressure (in atm) of propane in a mixture that contains equal weights of propane (C3H8) and butane (C4H10) at 18°C and 746 mmHg

  • chemistry -

    Ppropane = Xpropane*Ptotal
    Ptotal = 746/760 = ?atm (the problem asks for atm).

    So all we need to do is to determine Xpropane and plug into the above.
    molar mass C3H8 = about 44
    molar mass C4H10 = about 58. YOu can do all of this more accurately.
    Xpropane = (g/44)/[(g/44)+(g/58)]
    Solve for X and substitute into the above. I get something like 420 mm (but the problem asks for atm).

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