Chemistry help?

Question

Ethylene glycol (C₂H₆O₂) is used as an additive to the water in your automobile to lower its freezing point. A solution of ethylene glycol in water has a freezing point of -5.80 °C. What mass (in grams) of ethylene glycol must be added to 1000 g of this solution to lower the freezing point to -11.00 °C? (Kf for water is 1.86 °C･kg/mol).

Answer 1

The equation for the freezing point depression that will be used is...

∆T = imK

∆T = change in freezing point = 11.00º - 5.80º = 5.20º

i = van't Hoff factor = 1 for ethylene glycol (a non electrolyte)

m = molality = moles ethylene glycol / kg solvent

K = 1.86º/m

Solving for m (moles of ethylene glycol / kg ) we have...

m = ∆T / iK = 5.20º / (1)(1.86º/m)

m = 2.796 moles / kg

Since we have 1000 g of solution (i.e., 1 kg), we will need to add 2.796 moles of ethylene glycol to the existing solution to lower the freezing point to -11.00º.

To convert this to grams, we use the molar mass of ethylene glycol (C₂H₆O₂) of 62.07 g / mole:

2.796 moles x 62.07 g / mol = 174 g ethylene glycol need to be added