An ethylene glycol solution contains 25.4 g of ethylene glycol (C2H6O2) in 89.0 mL of water. (Assume a density of 1.00 g/mL for water.) You may want to reference (Pages 528 - 538) section 14.6 when completing this problem. Part A Determine the freezing point of the solution.

-7.44°C

Explanation:

Calculate the molality of the solution. Use the density of the solvent(water) as a conversion factor in order to convert from millilitres of solvent to grams of solvent. Then convert grams into kilograms. Finally, use the molar mass of ethylene glycol as a conversion factor to convert from grams to moles of ethylene glycol.  

m = 25.4 g C2H6O2/89.0 mL solv

   = 4.6321 C10H8O

Compute the freezing-point depression.  

ΔT_f=K_f*m ==> (1.86°C)*(3.9996 m)

       =7.44°C

Compute the freezing point of the solution by subtracting the freezing-point depression to the freezing point of the pure solvent.  

 freezing point =0.0°C-ΔT_f

                        = -7.44°C