Question

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform, CHCl3, is 173.1 mm Hg at 25°C. In a laboratory experiment, students synthesized a new compound and found that when 14.30 grams of the compound were dissolved in 246.0 grams of chloroform, the vapor pressure of the solution was 168.8 mm Hg. The compound was also found to be nonvolatile and a nonelectrolyte. What is the molecular weight of this compound? chloroform = CHCl3 = 119.4 g/mol MW= g/mol

Answer 1

According to Raoult’s law:
P = Po*X(solvent)
168.8 = 173.1*X(solvent)
X(solvent) = 0.9752
This is mole fraction of CHCl3

Molar mass of CHCl3 = 119.4 g/mol


mass(CHCl3)= 246.0 g

use:
number of mol of CHCl3,
n = mass of CHCl3/molar mass of CHCl3
=(2.46*10^2 g)/(119.4 g/mol)
= 2.061 mol

X(CHCl3) = n(CHCl3)/( n(CHCl3) + n(solute))
0.9752 = 2.061 / ( 2.061+n(solute))
2.01+0.9752*n(solute) = 2.061
0.9752*n(solute) = 5.119*10^-2
n(solute) = 5.25*10^-2 mol

mass(solute)= 14.30 g


use:
number of mol = mass / molar mass
5.25*10^-2 mol = (14.3 g)/molar mass
molar mass = 272.4 g/mol
Answer: 272.4 g/mol