125ch13f(1) - 56 13.77) a) b) 57 13.80) 58 13.83) 59 13.90)...

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Unformatted text preview: 56 13.77) a) b) 57 13.80) 58 13.83) 59 13.90) 60 13.90) (cont.) 61 13.92) Note: To determine the moles of C2H5OH we could also use the fact that the ratio of mole fractions of two substances is the same as their ratio of moles. XC2H5OH mol C2H5OH ------------ = -----------------XC24H50 mol C24H50 XC2H5OH mol C2H5OH = ------------ (mol C24H50) XC24H50 = (0.08/0.92) (1.83 x 103 mol) = 159.13 mol C2H5OH 62 13.94) 63 13.95) 64 13.95) (cont.) ...
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This note was uploaded on 02/19/2012 for the course CHEM 125 taught by Professor Zellmer during the Winter '08 term at Ohio State.

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