Answer to Question #164467 in Chemistry for Phyroe

Question

If 8 liters of ethylene glycol, C2H4(OH) density = 1.113 g/ml, is placed in an automobile radiator and diluted with 32 liters of water, what is the approximate freezing point of the solution in degrees fahrenheit?

Answer 1

∆T = K_F·m

K_F is the molal freezing point depression constant (for water K_F=1.86)

m is the molality of the solute (moles of solute divided by the kilograms of solvent).

density = m / V

m = density x V

m (C₂H₄(OH)₂) = 1.113 x 8000 = 8904 g

n = m/M

M (C₂H₄(OH)₂) = 62 g/mol

n (C₂H₄(OH)₂) = 8904/62 = 143.6 mol

Water density is 1 kg/l. Therefore, 32 litres of water is equal to 32 kg.

Molality m (C₂H₄(OH)₂) = 143.6/32 = 4.5 m

∆T = 1.86 · 4.5 = 8.4°

Approximate freezing point of the solution in degrees of Celcius: 0 - 8.4 = -8.4°C (Fahrenheit 17.6°).