what would be the boiling point of this solution in °C? (Kb for benzene is 2.53 °C/m and the density of benzene is 0.877 g/mL)

Question

0.0711 kg of biphenyl (C₁₂H₁₀) is dissolve in benzene (C₆H₆) to create a solution with a total volume of 350.0 mL. (Assume the change in volume is negligible) If the boiling point of pure benzene is 80.1 °C, then what would be the boiling point of this solution in °C? (Kb for benzene is 2.53 °C/m and the density of benzene is 0.877 g/mL)

Answer 1

∆T = imK

∆T = change in boiling point temperature = ?

i = van't Hoff factor = 1 for biphenyl since it is a non electrolyte

m = molality = moles of biphenyl / kg benzene (biphenyl is the solute; benzene is the solvent) = see below

K = boiling constant = 2.53

First, we must find the molality, which is moles of solute / kg of solvent

moles solute = 0.0711 kg C12H10 x 1000 g / kg x 1mole C12H10 / 154.2 g = 0.4611 mols C12H10

kg solute = 350.0 ml benzene x 0.877 g / ml x 1 kg / 1000 g = 0.3070 kg solute

molality = m = 0.4611 mols / 0.3070 kg = 1.502 m

Now, going back to our original equation ∆T = imK and solving for ∆T, we have ...

∆T = (1)(1.502)(2.53)

∆T = 3.800º NOTE: this is the CHANGE in boiling point, and solutes INCREASE boiling point, so ...

New boiling point of the solution = 80.1º + 3.80º = 83.9ºC