General chemistry freezing point

Question

Camphor boils at 207.4 Celsius. What mass of ethanol (46.07g/mol) dissolved in 675.0 g of camphor will raise the boiling point to 211.6 Celsius?

Formula: delta Tb = Kb(constant boiling point elevation) x i( vant Hoff constant) x m( molality)

notes: The Kf wasn’t given to me and I know that I need to find the molality to help assist me to find the mass of ethanol that’s needed to raise the boiling point)

Answer 1

Kb for camphor = 5.95

∆T = imKb

∆T = 211.6 - 207.4 = 4.2º

i = van't Hoff factor =1 for ethanol, a non electrolyte

m = molality = moles ethanol/kg camphor = ?

Kb = 5.95

Solving for m ...

m = ∆T / (i)(K) = 4.2 / (1)(5.95)

m = 0.706 mols / kg

Since there are 675.0 g of camphor, that is = 0.675 kg of camphor

moles ethanol = 0.706 mols / kg x 0.675 kg = 0.476 mols ethanol

Mass of ethanol = 0.476 mols x 46.07 g / mol = 21.95 g ethanol