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

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

Step-by-step explanation:

The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.

`(p_o-p_s)/(p_o)=(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=(216.7 g)/(78.12 g/mol)=2.7739 mol`

`(73.03 mmHg-71.61 mmHg)/(73.03 mmHg)=(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.