Antifreeze protects a car from freezing and from overheating. Calculate the freezing-point depression of a solution containing exactly 100. g of ethylene glycol (C2H6O2) antifreeze in 0.500 kg of water. Kf of water is: -1.86 C/m.
I got => ∆T(f) = -6.00ᵒC, but if you want support on learning the subject you need to show some effort on the problem.
I see a almost -6C also.
while antifreeze protects a car from freezing (as it name implies) it also protects it from overheating. calculate the freezing point of a solution 200g ethylene glycol ( C2H6O2) antifreeze in 750 g water. Kf for water is 1.86°C/m
Ah, freezing-point depression! It's like when your car sees a scary movie and gets all chilled out. Let's calculate how much it cools down, shall we?
First, we need to find the moles of ethylene glycol (C2H6O2). The molar mass of C2H6O2 is 62.07 g/mol, so dividing the mass of the ethylene glycol (100 g) by the molar mass gives us approximately 1.609 moles.
Next, let's find the moles of water in the solution. The molar mass of water (H2O) is 18.02 g/mol. Dividing the mass of water (0.500 kg) by the molar mass gives us approximately 27.764 moles.
Now, we divide the moles of ethylene glycol by the moles of water to get the molality of the solution. 1.609 moles / 27.764 moles = 0.058 mol/kg.
Finally, we can use the formula for freezing-point depression:
ΔT = Kf * m
where ΔT is the freezing-point depression, Kf is the freezing-point depression constant (-1.86 °C/m), and m is the molality of the solution (0.058 mol/kg).
Substituting in these values, we get:
ΔT = -1.86 °C/m * 0.058 mol/kg
And the calculation gives us approximately -0.108 °C.
So, the freezing-point depression of the solution is approximately -0.108 °C. Now your car can laugh in the face of freezing temperatures! Or maybe it can just chill and enjoy a nice cup of antifreeze. Cheers!
To calculate the freezing-point depression of a solution, we need to use the formula:
ΔTf = Kf * m * i
Where:
ΔTf = freezing-point depression (in degrees Celsius)
Kf = molal freezing-point depression constant (in degrees Celsius per molal)
m = molality of the solution (in mol/kg)
i = van't Hoff factor (number of particles the solute dissociates into)
In this case, we have ethylene glycol (C2H6O2) as the solute and water as the solvent.
First, we need to calculate the molality (m) of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
Step 1: Calculate the number of moles of ethylene glycol.
Given:
Mass of ethylene glycol (C2H6O2) = 100 g
Molar mass of ethylene glycol (C2H6O2) = 2*12.01 + 6*1.01 + 2*16.00 = 62.07 g/mol
Number of moles = mass / molar mass
Number of moles of ethylene glycol (C2H6O2) = 100 g / 62.07 g/mol
Step 2: Convert the mass of water to kilograms.
Given:
Mass of water = 0.500 kg
Step 3: Calculate the molality (m).
Molality (m) = moles of solute / mass of solvent (in kg)
Now that we have the molality (m), we can calculate the freezing-point depression (ΔTf) using the formula mentioned earlier.
Given:
Kf (freezing-point depression constant for water) = -1.86 °C/m
Now we can substitute the values into the formula.
ΔTf = Kf * m * i
Since ethylene glycol does not dissociate in water, i = 1.
ΔTf = -1.86 °C/m * m * 1
Replace m with the calculated value from earlier.
Finally, calculate the freezing-point depression (ΔTf).