Antifreeze is ethylene glycol, C 2 H 4 (OH) 2 and is made from ethylene which is

Question

Antifreeze is ethylene glycol, C2H4(OH)2 and is made from ethylene which is produced from oil.  It is miscible in water (i.e., dissolves in any proportion), does not dissociate in water, is toxic to drink (especially to dogs who like the sweet taste), but fortunately is biodegradable and at normal automobile operating temperatures, is not flammable.  It can prevent freezing of the automobile cooling system down to -34oF and the advertisement says “winter - summer automobile coolant” — it also prevents boiling over in the summertime. For best protection, the directions read, “use a 50:50 mixture of ethylene glycol and water”. Wouldn't pure ethylene glycol do a better job of preventing freeze-up of your cooling system????  (Hint: the Fp of pure ethylene glycol is -11.5oC and the molal freezing point constant of water is 1.86 oC/molality)

What is the freezing point temperature in oC of water if a 46.79 percent mixture (by mass) of ethylene glycol in water is used? Put your answer in 3 significant figures.

Explanation / Answer

Ans. Step 1: Given, concentration of ethylene glycol (EG) = 46.79 % (w/w), that is, there is 46.79 g of solute (ethylene glycol) per 100.0 g of solution.

So, on a 100.0 g solution basis-

            Mass of EG = 46.79 g

                        Moles of EG = Mass / MW = 46.79 g / (62.07 g/mol) = 0.7538 mol

            Mass of solvent (water) = Mass of solution – Mass of EG

                                                = 100.0 g – 46.79 g

                                                = 53.21 g

                                                = 0.05321 kg

# Now, molality of solution = Moles of EG / Mass of solvent in kg

                                                = 0.7538 mol / 0.05321 kg

                                                = 14.1665 mol/ kg

                                                = 14.1665 m

# Step 2: Depression in freezing point of the solution is given by-

                        dTf = i Kf m             - equation 1

            where, i = Van’t Hoff factor. [ i = 1, assuming the solute be non-electrolyte]

                        Kf = molal freezing point depression constant (of water) = 1.860C / m

                        m = molality of the solution

                        dTf = Freezing point of pure solvent – Freezing point of solution

Putting the values in equation 1-

            dTf = 1 x (1.86 0C m-1) x 14.1665 m = 26.350C

Now,

            Freezing point of solution = dTf + Freezing point of pure solvent

                                                            = (-11.50C) – (26.350C)

                                                            = -37.850C

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