Answer:
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
Explanation:
The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.
[tex]\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}[/tex]
Where:
[tex]p_o[/tex] = Vapor pressure of pure solvent
[tex]p_s[/tex] = Vapor pressure of the solution
[tex]n_1[/tex] = Number of moles of solvent
[tex]n_2[/tex] = Number of moles of solute
[tex]p_o = 73.03 mmHg[/tex]
[tex]p_s= 71.61 mmHg[/tex]
[tex]n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol[/tex]
[tex]\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}[/tex]
[tex]n_2=0.05499 mol[/tex]
Mass of 0.05499 moles of estrogen :
= 0.05499 mol × 272.4 g/mol = 14.9802 g
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
