Question

The vapor pressure of ethanol is 54.68 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 203.1 grams of ethanol to reduce the vapor pressure to 53.09 mm Hg ? ethanol = CH3CH2OH = 46.07 g/mol.

Answer 1

According to Raoult’s law:

P = Po*X(solvent)

53.09 = 54.68*X(solvent)

X(solvent) = 0.9709

This is mole fraction of C2H5OH

mass of C2H5OH = 203.1 g

we have below equation to be used:

number of mol of C2H5OH,

n = mass of C2H5OH/molar mass of C2H5OH

=(203.1 g)/(46.07 g/mol)

= 4.409 mol

X(C2H5OH) = n(C2H5OH)/( n(C2H5OH) + n(solute))

0.9709 = 4.409 / ( 4.409+n(solute))

4.28+0.9709*n(solute) = 4.409

0.9709*n(solute) = 0.1282

n(solute) = 0.132 mol

Molar mass of estrogen = 272.4 g/mol

we have below equation to be used:

mass of estrogen,

m = number of mol * molar mass

= 0.132 mol * 272.4 g/mol

= 35.96 g

Answer: 35.96 g