Question

A 2.000g sample of a biomolecule was dissolved in 15.0g of carbon tetrachloride. This solution boiled at 77.85 degrees Celsius. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling point constant is 5.03 degrees Celsius kg/mol, and the boiling point of pure carbon tetrachloride is 76.50 degrees Celsius.

Answer 1

Reaction solution temeprature (T_solution) is set to 77.85 C instead of 76.50 C (T_solvent) to make the vapor pressure of CCl4 equal to external pressure. This is boiling point elevation. Change in boiling temperature

  T_solution - T_solvent = 77.85 - 76.50 = 1.35 ⁰C

Boiling point constant K_b of CCl4 = 5.03 ⁰C.Kg/mol

Then find out the molality (m) of the solution

m = 1.35 ^oC / 5.03 ⁰C.Kg/mol

m = 0.268 mol/kg

Molality is the number moles of solute dissolved in a kg of solvent.

CCl4 amount is 15.0 grams = 0.015 kg

m = moles of solute / kg solvent

0.268 mol/kg X 0.015 kg = moles of solute

= 0.00402 moles

Number of moles = weight / molecular weight

Weight of sample is 2.000 grams

0.00402 mols = 2 grams / molecular weight

molecular weight = 2 grams / 0.00402 mols

= 497.512 grams/mol

So the molar mass of sample is 497.512 grams/mol

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