Question

The vapor pressure of ethanol is 54.68 mm Hg at 25C. How many grams of testosterone, C19H28O2, a nonelectrolyte (MW=288.4g/mol), must be added to 270.0 grams of ethanol to ethanol =CH3CH2OH=46.07 g/mol. _________ g. Testosterone

Answer 1

Step 1

We will use the colligative property relative to the lowering of vapor pressure. The formula:

ΔP = Psv° . X

ΔP = Pressure sv pure - Pressure of the solution

X means molar fraction for solute

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Step 2

The vapor pressure of the solution at 25°C is 54.68 mmHg and:

54,68 mmHg - P sl = 54,68 mmHg . X

P sl = 53.64 mmHg (please read the question)

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Step 3

X is calculated first:

54.68 mmHg - P sl = 54.68mmHg . X

54.68 mmHg - 53.64 mmg = 1.04 mmHg = 54.68 mmHg . X

=> X = 0.0109 (solute) = moles from solute / moles from solute + moles from solvent (1)

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Step 4

Moles of ethanol = mass of ethanol/molar mass of ethanol

Moles for ethanol: 270 g / 46.07 g/m = 5.86 moles of ethanol

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Step 5

Moles from solute are cleared from (1) and the result is:

Moles from solute = 0.0109 x moles from solvent/(1-0.0109) =

Moles from solute = 0.0109 x 5.86 moles/(1-0.0109) = 0.065 moles

Moles from solute = mass of solute/molar mass solute = 0.065 moles

Mass of solute = 0.065 moles x 288.4 g/mol = 18.7 g of solute

Answer: 17.7 g Testosterone