Question

using Raoult's Law

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 281.0 grams of benzene to reduce the vapor pressure to 72.01 mm Hg ?

benzene = C₆H₆ = 78.12 g/mol.

______ g estrogen

Answer 1

According to Raoult's law ,we have relation P ⁰ - P _s / P _s = W ₂ M ₁ / W ₁ M ₂ -----------> ( 1)

Where , P ⁰ is a V.P of pure solvent , P _s is V.P of solution , W ₂ is a mass of non volatile solute , M ₂ is molar mass of non volatile solute , W ₁ is mass of solvent and M ₁ is molar mass of solvent.

Given

V.P of benzene ( P ⁰ ) = 73.03 mm Hg

molar mass of benzene ( M ₁ ) = C6H6 = ( 6 12.01 ) + ( 6 1.0079 ) = 78.11 g / mol

Mass of benzene ( W ₁ ) = 281.0 g

V.P of solution ( P _s ) = 72.01 mm Hg

Molar mass of solute ( estrogen ) = 272.4 g /mol

Mass of estrogen = ?

Substituting these values in above relation 1 , we get

73.03 - 72.01 / 72.01 = W ₂ 78.11 / 281.0 272.4

1.02/ 72.01 = W ₂ 78.11 / 765444

0.01416 = W ₂ 78.11 / 765444

W ₂ 78.11 = 0.01416 765444

W ₂ 78.11 = 10838.7

W ₂ = 10838.7 / 78.11

W ₂ = 138.76 g

ANSWER : 138.8 g estrogen