The vapor pressure of ethanol is 54.68 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 239.0 grams of ethanol to reduce the vapor pressure to 54.11 mm Hg ?

The mass of estrogen that must be added is 2.83 grams

Explanation:

The equation used to calculate relative lowering of vapor pressure follows:

where,

= relative lowering in vapor pressure

i = Van't Hoff factor = 1 (for non electrolytes)

= mole fraction of solute = ?

= vapor pressure of pure ethanol = 54.68 mmHg

= vapor pressure of solution = 54.11 mmHg

Putting values in above equation, we get:

This means that 0.0104 moles of estrogen are present in the solution

To calculate the number of moles, we use the equation:

Moles of estrogen = 0.0104 moles

Molar mass of estrogen = 272.4 g/mol

Putting values in above equation, we get:

Hence, the mass of estrogen that must be added is 2.83 grams

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