How much ethylene glycol must be added to 1.1 L of water to keep it from freezing at -13 degreeF? The Kf value is 1.86 degree C/molal?
How much ethylene glycol must be added to 1.1 L of water to keep it from freezing at -13 degreeF ? The Kf value is 1.86 degree C/molal. Explain how you determined your answers.
- 1 month ago
There are three important factors when calculating freezing point depression:
(1) The freezing point depression constant, kf, which is given
(2) The molality of the solution, which we're solving for
(3) the Van't Hoff factor, which is the number of particles formed when the compound we add dissolves in an ideal solution
Ethylene glycol is very weakly acidic, but its dissociation in the water is negligible so we'll use a Van't Hoff factor of i = 1.
Finally, we need to convert our -13 degrees Fahrenheit to Celsius, which gives us -25 degrees Celsius. Water normally freezes at 0 degrees Celsius so the freezing point depression is 25 degrees Celsius.
So now we have everything we need to use our freezing point depression formula:
delta_T = kf * m * i
25 degrees Celsius = 1.86 C/m * molality * 1 => molality = 13.44 m (molal)
Molality is defined as moles of solute divided by mass of solvent, so next we need to find the mass of 1.1 L of water. You should have one of the forms of the density of water memorized, either 1 g/mL or 1000 kg/m^3, either one gives us 1.1 kg of water.
So the amount of ethylene glycol you would have to add would be 13.44 * 1.1 = 14.8 moles, or about 0.918 kg.
Hope this helps!