Question

The vapor pressure of benzene is 73.03 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 216.7 grams of benzene to reduce the vapor pressure to 71.61 mm Hg ? benzene = C6H6 = 78.12 g/mol.

Answer 1

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.

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

Where:

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of the solution

$n_1$ = Number of moles of solvent

$n_2$ = Number of moles of solute

$p_o = 73.03 mmHg$

$p_s= 71.61 mmHg$

$n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol$

$\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}$

$n_2=0.05499 mol$

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.