Calculate the boiling point of a solution made by dissolving 0.240 mol naphthalene in 2.45 mol CHCl₃.

Question

For CHCl₃, the normal boiling point is 61.2⁰C. Kb = 3.63 ⁰C/m. The molar mass of CHCl₃ = 119.4 g/mol.

Answer 1

∆T = imK

∆T = change in boiling point = ?

i = van't Hoff factor = 1 for naphthalene because it is a non electrolyte

m = molality = mols naphthalene / kg CHCl3 (see below for calculation of m)

K = boiling constant = 3.63ºC / m

calculation of m:

mols naphthalene = 0.240 mol

kg CHCl3 = 2.45 mol CHCl3 x 119.4 g / mol x 1 kg / 1000 g = 0.2925 kg

m = 0.240 mol / 0.2925 kg

m = 0.8205 mol / kg

∆T = imK

∆T = (1)(0.8205)(3.63)

∆T = 2.98º

New boiling point of the solution is the original temperature + 2.98º...

61.2º + 2.98º = 64.2º = boiling point of solution