Question

A solution prepared at 20 degrees celsius by dissolving 7.44g of an unknown hydrocarbon in 75g benzene (C6H6, 78.112 g/mol, 0.8765 g/cm^3) is found to have a vapor pressure of 376 mmHg at 60 degrees celsius. The vapor pressure of pure benzene at 60 degrees celsius is 400 mmHg. Determine the molar mass of the unknown compound.

Answer 1

Solution :-

Mass of unknown hydrocarbon = 7.44 g

Mass of benzene = 75 g

Molar mass of benzene = 78.112 g per mol

Vapor pressure of benzene = 400 mmHg

Vapor pressure of solution = 376 mmHg

Molar mass of unknown hydrocarbon =?

Using the Raoults law formula we can calculate the mole fraction of the benzene.

P_solution = (χ_solvent) (P°_solvent)

Where ,   P_solution = vapor pressure of solution , (χ_solvent) = mole fraction of solvent , = vapor pressure of pure solvent

Now lets put the values in the formula

376 mmHg = (χ_solvent) * 400 mmHg

376 mmHg / 400 mmHg = (χ_solvent)

0.94 = (χ_solvent)

Therefore mole fraction of the solvent = 0.94

Mole fraction of the solute = 1- 0.94 = 0.06

Now lets calculate the moles of benzene using its given mass

Moles = mass / molar mass

Moles of Benezene =75 g /78.112 g per mol = 0.96016 mol benzene

Now use the mole of the benzene and mole fraction of benzene to determine the moles of the unknown hydrocarbon

0.94 = 0.96016 / (0.96016+x)

X is nothing but the moles of the unknown hydrocarbon

0.94 * 0.96016+x =0.96016

0.90255 +0.94 x =0.96016

0.94 x = 0.05851

X = 0.05851/0.94

X =0.062244 mol unknown hydrocarbon

Now lets use this moles of the unknown hydrocarbon to calculate its molar mass

Molar mass = mass / moles

Molar mass of unknown hydrocarbon = 7.44 g / 0.062244 mol

                                                             = 119.53 g per mol

Therefore molar mass of the unknown hydrocarbon = 119.53 g / mol