What is the molar mass of the hydrocarbon?

Question

A solution contains 3.75 g of a nonvolatile pure hydrocarbon in 95 g acetone, (CH₃)₂CO. The boiling points of pure acetone and the solution are 55.95 °C and 56.50 °C, respectively. The molal boiling-point constant of acetone is 1.71 °C/m. What is the molar mass of the hydrocarbon?

Answer 1

∆T = imK

∆T = change in boiling point = 56.50 - 55.95 = 0.55^o

i = van't Hoff factor = 1 for a hydrocarbon (a non electrolyte)

m = molality = mols hydrocarbon / kg acetone = ?

K = boiling constant = 1.71º/m

Solving for m, we have...

m = ∆T / (i)(K) = 0.55 / (1)(1.71) = 0.3216 moles / kg acetone

Since we have 95 g acetone, that is = 0.095 kg acetone, so ...

moles of hydrocarbon = 0.3216 mols / kg x 0.095 kg = 0.0306 mols hydrocarbon

molar mass of hydrocarbon = g / moles = 3.75 g / 0.0306 mols = 123 g / mol (3 sig. figs.)

Answer 2

The boiling point elevation = 56.50 - 55.95 = 0.55^oC

The molar mass of the hydrocarbon is calculated

M = (1000 x w2 x kb) / (boiling point elevation x w1)

where w2 is the mass of solute added = 3.75 g

kb = 1.71^oC/m

w1 = mass of solvent = 95 g

So putting all the values :

M = (1000 x 3.75 x 1.71) / (0.55 x 95)

= 122.73 g/mol