An antifreeze solution is prepared containing 40.0 g of ethylene glycol (molar mass = 62.0 g/mol) in 60.0 g of water. Calculate the freezing point of this solution. Which calculation is it?
A. 120.1⁰C
B. -20.1⁰C
C. 0.518 ⁰C
D. 20.1⁰C
LOL, hint - it is below freezing point of water with no antifreeze
Thanks :)
You are welcome ;)
To calculate the freezing point of the solution, we need to use the equation for freezing point depression.
The freezing point depression is given by the equation:
∆T = Kf * molality
Where:
∆T is the change in freezing point
Kf is the cryoscopic constant (freezing point depression constant) of the solvent (water in this case)
molality is the number of moles of solute per kilogram of solvent
First, let's calculate the number of moles of ethylene glycol (solute):
moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol
moles of ethylene glycol = 40.0 g / 62.0 g/mol
moles of ethylene glycol = 0.645 mol
Next, let's calculate the molality of the solution:
molality = moles of solute / Mass of solvent (in kg)
molality = 0.645 mol / 0.0600 kg
molality = 10.8 mol/kg
Now, using the cryoscopic constant for water (Kf = 1.86 °C/m), we can calculate the change in freezing point (∆T):
∆T = Kf * molality
∆T = 1.86 °C/m * 10.8 mol/kg
∆T ≈ 20.1 °C
The freezing point of the solution is given by the sum of the freezing point of the pure solvent (water) and the depression (∆T):
Freezing point of solution = Freezing point of water - ∆T
Freezing point of solution = 0 °C - 20.1 °C
Freezing point of solution ≈ -20.1 °C
Therefore, the correct answer is B. -20.1 °C.