Question

compound with the empirical formula C8H5 is dissolved in benzene at 25°C. When 10.0 g of the compound are dissolved in 100 g of benzene, the vapor pressure above the resulting solution is 91.64 mm Hg. If the vapor pressure of pure benzene is 95.18 mm Hg at 25°C, calculate the molecular weight and determine the molecular formula of the compound. You may assume the compound is nonvolatile and that benzene obeys Raoult’s law in the solution.

Answer 1

The lowering of vapour pressure is a colligative property

Vapour pressure of solution = vapour pressure of pure solvent X mole fraction of solvent

91.64 = 95.18 X mole fraction of benzene

0.962 = mole fraction of benzene

The mass of benzene = 100grams

Mol wt of benzene = 78 g

So moles of benzene = mass / mol wt = 100 / 78 = 1.28 moles

Mole fraction = moles of benzene / moles of benzene + moles of solute

0.962 = 1.28 / 1.28 + moles of solute

1.23 + moles of solute X 0.962 = 1.28

Moles of solute = 0.05 / 0.962 = 0.052 moles

molecular weight of given compound = mass of compound / mol wt of compound

0.052 = 10 / mol wt

mol wt of compound = 192.30

The empirical formula of the compound = C8H5

Let the molecular formula of the compound is C8xH5x

so

12 X 8 X x + 5x = 192.30

101x = 192.3

x = 1.9

So the probalble molecular formula will be

C16H10