chemistry
posted by Anonymous .
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

P_{propane} = X_{propane}*P_{total}
P_{total = 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).}