Question

At 25^o C the vapor pressure of pure benzene is 0.1252 atm. If we dissolve 9 g of an unknown hydrocarbon in 88.9 g of pure benzene (MW = 78.0 g/mol), we observe that the vapor pressure of the benzene in the resulting solution is 0.1191 atm. What is the molecular weight of the unknown solid (in g/mol)?

Answer 1

Let the mol. wt. of the unknown be M g/mol.

therefore no. of moles of unknown in the mixture = 9/ M moles

and no. of moles of benzene = 88.9/ 78.0 = 1.14 moles

Therefore mole fraction of the unknown = (9/M) / (9/M) + 1.14

Now as per Raoult's law the relative lowering of the vapor pressure of the solvent is equal to the mole fraction of the solute

Hence, (0.1252 - 0.1191)/ 0.1252 = (9/M) / (9/M) + 1.14

or 0.0061/ 0.1252 = (9/M) / (9/M) + 1.14

or 0.050 = (9/M) / (9/M) + 1.14

or (0.45/M) + 0.057 = 9/M

or 8.55/M = 0.057

or M = 8.55/ 0.057 = 150 g/ mol