Question

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 228.0 grams of ethanol to reduce the vapor pressure to 53.46 mm Hg ? ethanol = CH3CH2OH = 46.07 g/mol.

Answer 1

Answer:

30.5 g of estrogen

Explanation:

Lowering vapor pressure formula → P° - P' = P° . Xm

P° → Vapor pressure of pure solvent → 54.68 mmHg

P' → Vapor pressure of solution → 53.46 mmHg

54.68 mmHg - 53.46 mmHg = 54.68 mmHg . Xm

0.0223 = Xm → This is the mole fraction for solute

Moles of solute / Total moles

Total moles = Moles of solute + Moles of solvent

Let's determine the moles of solvent → 228 g / 46.07 g/mol = 4.95 moles

Let's determine the moles of solute

Moles of solute / Moles of solute + 4.95 = 0.0223

Moles of solute = 0.0223 . Moles of solute + 0.110

0.9777 moles of solute = 0.110

Moles of solute = 0.110 / 0.9777 → 0.112

Now we can convert the moles to mass to finish the excersise:

0.112 mol . 272.4 g/mol = 30.5 g