How many grams of testosterone, C$_{19}$H$_{28}$O$_2$, a nonvolatile, nonelectrolyte (MW = 288.4 g/mol), must be added to 233.2 grams of ethanol to reduce the vapor pressure to 54.11 mm Hg? ethano = CH$_3$CH$_2$OH = 46.07 g/mol. Mass = g
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Moles of ethanol = 233.2 g / 46.07 g/mol = 5.06 mol Moles of testosterone = x g / 288.4 g/mol = x / 288.4 mol Total moles in the solution = 5.06 + x / 288.4 Mole fraction of ethanol = 5.06 / (5.06 + x / 288.4) Show more…
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