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.

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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explanation:

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i