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24 September, 10:53

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.

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  1. 24 September, 11:20
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    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
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