Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C).

Question

. Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C). Calculate (a) the freezing point and (b) the boiling point of a solution containing 685 g of ethylene glycol in 2075 g of water. (For water Kf= 1.86 °C/m and Kb=0.52 °C/m)

Answer 1

ΔT_f = i K_f m

where ΔT_f is the freezing point lowering, K_f is the freezing point constant, and m is the molality.

i (van't Hoff factor) = 1 since ethylene glycol is a non-electrolyte.

T_f of pure water (T_f°) = 0.00 °C

T_f of solution = ? °C

K_f = 1.86 °C/m

molality m = moles of solute / kg of solvent = moles of ethylene glycol / kg of water

mol ethylene glycol = 685 g x (1 mol / 62.07 g) = 11.0 mol ethylene glycol

kg of water = 2075 g x (1 kg / 1000 g) = 2.075 kg of water

molality = 11.0 mol / 2.075 kg = 5.30 m

ΔT_f = K_f m

ΔT_f = 1.86 °C/m x 5.30 m = 9.86 °C

T_f° - T_f = 9.86 °C

0.00 - T_f = 9.86 °C

T_f = 0.00 - 9.86 = -9.86 °C

Similarly, ΔT_b = K_b m

ΔT_b = 0.52 °C/m x 5.30 m = 2.76 °C

T_b - T_b° = 2.76 °C

T_b - 100.00 = 2.76 °C

T_b = 100.00 + 2.76 = 102.76 °C

Answer 2

Delta T = imK

Delta T = change in temp

i = van’t Hoff factor = 1 for ethylene glycol

m = molality = 685 g x 1 mole/62.1 g / 2.075 kg = 5.32 molal

K = 1.86

Delta T = (1)(5.32)(1.86) = 9.89 degrees

New f.p. = 9.89 degrees C below the normal freezing point.

For b.p., do the same thing using K = 0.52.