Question

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of testosterone, C19H28O2, a nonvolatile, nonelectrolyte (MW = 288.4 g/mol), must be added to 198.3 grams of diethyl ether to reduce the vapor pressure to 457.31 mm Hg ? diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol. g testosterone

Answer 1

mass testosterone = 10.56 g

Explanation

Vapor pressure of solution = 457.31 mmHg

vapor pressure of pure diethyl ether = 463.57 mmHg

vapor pressure lowering = (vapor pressure of pure diethyl ether) - (Vapor pressure of solution)

vapor pressure lowering = (463.57 mmHg) - (457.31 mmHg)

vapor pressure lowering = 6.26 mmHg

mole fraction of testosterone = (vapor pressure lowering) / (vapor pressure of pure diethyl ether)

mole fraction of testosterone = (6.26 mmHg) / (463.57 mmHg)

mole fraction of testosterone = 0.0135

mole fraction of testosterone = (moles testosterone) / (moles testosterone + moles diethyl ether)

0.0135 = (X) / (X + (198.3 g / 74.12 g/mol))

Solving for X, X = 0.0366 mol

moles testosterone = X = 0.0366 mol

mass testosterone = (moles testosterone) * (molar mass testosterone)

mass testosterone = (0.0366 mol) * (288.4 g/mol)

mass testosterone = 10.56 g