A 115 mL sample of a 10.0 M ethylene glycol (C2H6O2) solution is diluted to 1.30 L. What is the freezing point of the final solution? (Assume a density of 1.06 g/mL for the final solution.) What is the boiling point of the final solution?

Question

How would I go about setting this up? thanks!

Answer 1

10.0 M x (115 mL/1300) = 0.885 M.
That's 0.883 mols/L soln.
1000 mL x 1.06 g/mL = 1060g soln is the mass of 1000 mL soln.
0.883 mols ethylene glycol x molar mass = 0.883 x 62.07 = 54.8 g is the mass of ethylene glycol in a liter (1060 g) of soln.
1060 - 54.8 = 1005.2 g is the solvent or 1.005 kg for the solvent and 0.883 mols solute or 0.883/1.005 = molality ethylene glycol.

Then delta T = Kf*m and subtract from zero to obtain freezing point.
delta T = Kb*m and add to 100 to obtain boiling point.