Question

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform , CHCl₃, is 173.11 mm Hg at 25 °C.

In a laboratory experiment, students synthesized a new compound and found that when 14.88 grams of the compound were dissolved in 230.8 grams of chloroform, the vapor pressure of the solution was 168.61 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight of this compound?

chloroform = CHCl₃ = 119.40 g/mol.

NOTE : This proplem related to  Raoult's Law

Answer 1

m = 14.88 sample

m = 230.8 g of Chloroform

P° = 168.61 mm Hg

since it is nonvolatile then

dP = xsolute * P°solvent

Pmix = P°solvent - dP

then

substitute all values

Pmix = P°solvent - xsolute * P°solvent

Pmix = (1-xsolute) * P°solvent

168.61 = (1-x) * (173.11)

x = -(168.61 /173.11-1)

x = 0.02599503206

x S= mol of S/ (mol of S + mol of solvent)

mol of solvnet = mas/MW = 230.8/119.4 = 1.9329983

0.02599503206 = (mol of S)/(mol of S + 1.9329983)

(mol of S + 1.9329983) = 1/(0.02599503206 ) * mol of S

(mol of S + 1.9329983) = 38.46 mol of S

-37.46 mol of S = -1.9329983

mol of S= 1.9329983/(37.46) =0.0516016631 mol

then

MW of sample = mass of sample / mol of sample = 14.88/0.0516016631 = 288.362 g /mol