How much antifreeze (ethylene glycol) must be added to one liter of water in order to lower the freezing point of water by 25 degrees c?

0.8342 grams of ethylene glycol must be added to one liter of water.

Explanation:

Volume of water or solvent = 1 L

Density of water = 1 g/L

Mass of the water = Density ×  Volume  = 1 g/L × 1 L = 1 g

1 g = 0.001 kg

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by ethylene glycol is 1 .

i = 1

Moles of solute (ethylene glycol) = 0.013440860 moles

Mass of 0.013440860 moles of ethylene glycol:

0.8342 grams of ethylene glycol must be added to one liter of water.

the answer is 0.8m.

by: nicholas