antifreeze used in automotive radiartors contains ethylene glycol C2H6O2. it is

Question

antifreeze used in automotive radiartors contains ethylene glycol C2H6O2. it is used to prevent freezing or boiling of the radiator fluid which is normally water. the normal freezing and boiling point temperature of water is 0C and 100C respectively.
A- if the weather dips to -10 C (14 F or 18 below zero). what mass of ethylene glycol need to be dissolved in 2.5kg of water to prevent freezing at this temperature?.
B- ehat is the boiling point for this same concentration of ethylene glycol solution?

Tf for water is 0C, Kf=1.86C/m
Tb for water = 100 C, Kb= 0.512 C/m
Ivh =1.

Explanation / Answer

Ans. A. Freezing point depression, dTf is given by-

            dTf = i Kf m                - equation 1

            where, i = Van’t Hoff factor = 1 for non-dissociating solutes

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

                        m = molality of the solution = m

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

                                    = 00C - (-100C) = 100C

Putting the values in above equation,       

            100C = 1 x (1.860C/m) x m

            Or, m = 100C / (1.860C/m) = 5.376 m

So, molality of desired solution = 5.376 m

Moles of ethylene glycol required for 2.5 kg solvent = molaity x mass of solvent

                                                = 5.376 m x 2.5 kg

                                                = (5.376 moles /kg) x 2.5 kg

                                                = 13.441 moles

Mass of ethylene glycol required = number of moles x molar mass

                                                = 13.441 moles x 62.068 g mol-1

                                                = 834.26 g

Ans. B. Boiling point elevation, dTb is given by-

            dTb = i Kb m               - equation 2

            where, i = Van’t Hoff factor = 1 for non-dissociating solutes

                        Kb = molal boiling point elevation constant (of the solvent) = 0.5120C/m

                        m = molality of the solution = 5.376 m

                        dTb = bp of solution – bp of pure solvent

Putting the values in above equation-

            dTb = 1 x (0.5120C/m) x 5.376 m = 2.750C

Now,

            dTb = bp of solution – bp of pure solvent

            or, 2.750C = bp of solution – bp of pure solvent

            or, 2.750C + bp of pure solvent = = bp of solution

            or, = bp of solution = 2.750C + 1000C = 102.750C

Thus, boiling point of the resultant solution = 102.750C

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