Question

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was determined to be 77.85c. calculate the molar mass of the biomolecule. for carbon tetrachloride, the boiling-point constant is 5.03c kg/mol, and the boiling point of pure carbon tetrachloride is 76.50c.

Answer 1

We will use boiling point formula:

ΔT = i Kb m 

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality 

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass  Kg

           0.27 = moles /  0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
                               = 2 g /  0.00405 moles 
                               = 493.8 g /mol