Question

The vapor pressure of chloroform is 173.11 mm Hg at 25°C. How many grams of testosterone, C19H28O2, a nonvolatile, nonelectrolyte (MW = 288.4 g/mol), must be added to 220.5 grams of chloroform to reduce the vapor pressure to 170.36 mm Hg ? chloroform = CHCl3 = 119.40 g/mol.

Answer 1

According to Raoult’s law:

P = Po*X(solvent)

170.36 = 173.11*X(solvent)

X(solvent) = 0.984114

This is mole fraction of chloroform

mass(chloroform)= 220.5 g

number of mol of chloroform,

n = mass of chloroform/molar mass of chloroform

=(220.5 g)/(119.4 g/mol)

= 1.847 mol

X(chloroform) = n(chloroform)/( n(chloroform) + n(C19H28O2))

0.9841 = 1.847 / ( 1.847+n(C19H28O2))

1.817+0.9841*n(C19H28O2) = 1.847

0.9841*n(C19H28O2) = 2.934*10^-2

n(C19H28O2) = 2.981*10^-2 mol

molar mass of C19H28O2 = 288.4 g/mol

so,

mass of C19H28O2 = molar mass of C19H28O2 * number of mol of C19H28O2

= 288.4 g/mol * 2.981*10^-2 mol

= 8.60 g

Answer: 8.60 g