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 240.8 grams of benzene to reduce the vapor pressure to 71.04 mm Hg ?

benzene = C6H6 = 78.12 g/mol.

________g estrogen

Answer 1

let x gm estrogen is added.

then moles of extrogen = x / 272.4 mole.

moles of solvent (benzene) = 240.8 / 78.12 = 3.08 mole.

mole fraction of estrogen(X2) = x / 272.4 / (x / 272.4 + 3.08)

we know,

P10 - P1 / P10 = X₂

(73.03 - 71.04) / 73.03 =  x / 272.4 / (x / 272.4 + 3.08)

0.027 =  x / 272.4 / (x / 272.4 + 3.08)

1.00 * 10^-4 x + 0.08316 = x / 272.4

1.00 * 10^-4 x + 0.08316 = 3.67 * 10^-3 x

x = 23.29 gm

so mass of estrogen added = 23.29 gm.