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
Please show work!!
dT = Kf*m
m = molality = mols/kg solvent
You have how many mols. mols = grams/molar mass = ?
Then m = mols/0.5
Then substitute Kf and m and solve for delta T.
Post your work if you get stuck.
To calculate the freezing-point depression of the solution, we can use the formula:
ΔTf = Kf * m * i
where:
ΔTf is the freezing-point depression
Kf is the freezing-point depression constant (in this case, it is -1.86 °C/m for water)
m is the molality of the solution (the amount of moles of solute per kilogram of solvent)
i is the Van't Hoff factor (the number of particles the solute dissociates into)
First, we need to determine the molality (m) of the solution, which is the moles of ethylene glycol per kilogram of water.
Molar mass of ethylene glycol (C2H6O2):
C: 12.01 g/mol
H: 1.01 g/mol (x6)
O: 16.00 g/mol (x2)
Molar mass of C2H6O2 = 2(12.01) + 6(1.01) + 2(16.00) = 62.07 g/mol
So, the number of moles of ethylene glycol (C2H6O2) in 100.0 g is:
moles = mass / molar mass = 100.0 g / 62.07 g/mol = 1.61 mol
Next, we calculate the molality (m):
m = moles of solute / mass of solvent in kg
m = 1.61 mol / (0.500 kg) = 3.22 mol/kg
Now, we can calculate the freezing-point depression (ΔTf):
ΔTf = Kf * m * i
Since ethylene glycol does not dissociate in water, the Van't Hoff factor (i) is 1.
ΔTf = (-1.86 °C/m) * (3.22 mol/kg) * 1 = -6.00 °C
Therefore, the freezing-point depression of the solution is -6.00 °C.
To calculate the freezing-point depression, we need to use the formula:
∆T = Kf * m * I
Where:
∆T is the freezing-point depression in degrees Celsius
Kf is the freezing-point depression constant for the solvent (-1.86 °C/m for water)
m is the molality of the solution in moles of solute per kilogram of solvent
I is the van't Hoff factor (which is 1 for ethylene glycol)
First, we need to convert the mass of ethylene glycol to moles. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol.
Moles of ethylene glycol = 100 g / 62.07 g/mol = 1.61 mol
Next, we calculate the molality (moles of solute per kilogram of solvent) using the mass of water:
Mass of water = 0.500 kg
Molality = moles of solute / mass of solvent (in kg)
Molality = 1.61 mol / 0.500 kg = 3.22 mol/kg
Finally, we can calculate the freezing-point depression (∆T):
∆T = -1.86 °C/m * 3.22 mol/kg * 1 = -6.00 °C
Therefore, the freezing-point depression of the solution is -6.00 °C.