A solution is prepared by adding 35.5 mL of ethylene glycol,C₂H₆O₂,to 250.0 g of water.What is the freezing point of this solution? The K_f for water is 1.86°C/m and the density of ethylene glycol is 1.1135 g/mL.Consider if the solute is an electrolyte or a nonelectrolyte.

Question

Quizplus · Accepted Answer

First, we need to determine if the solute (ethylene glycol) is an electrolyte or a nonelectrolyte. Since it is not specified in the problem, we will assume it is a nonelectrolyte.

Next, we can use the formula: 
ΔTf = Kf*m 
where ΔTf is the freezing point depression, Kf is the freezing point depression constant (1.86°C/m for water), and m is the molality of the solution.

To find the molality, we need to first convert the mass of water into moles: 
250.0 g H2O * (1 mol H2O/18.015 g H2O) = 13.876 mol H2O

Next, we can use the volume and density of ethylene glycol to find the mass: 
35.5 mL C2H6O2 * (1.1135 g C2H6O2/mL) = 39.505 g C2H6O2

Then, we can convert the mass of ethylene glycol into moles: 
39.505 g C2H6O2 * (1 mol C2H6O2/62.07 g C2H6O2) = 0.6359 mol C2H6O2

Finally, we can use the molality formula: 
m = moles of solute / mass of solvent (in kg) 
m = 0.6359 mol / 0.2500 kg = 2.544 m

Now we can plug this into the formula for ΔTf: 
ΔTf = 1.86°C/m * 2.544 m = -4.74°C

Since the freezing point depression is negative, the freezing point of the solution will be lower than the freezing point of pure water. To find the actual freezing point, we need to subtract the ΔTf from the freezing point of pure water: 
0°C - 4.74°C = -4.74°C

Therefore, the answer is B) -4.74°C.

A Solution Is Prepared by Adding 35