The freezing point of ethanol is -114.6 C and its Kf value is 2.00 C/m. What is the freezing point for the solution prepared by dissolving 50.0 g of glycerin (C3H8O3) in 200.0 g ethanol?
I used Change in temp=Kfm
kf=2.00
I calculated m to be 2.1736.
I ended up the answer of -118.4 C but the answer is apparently -120. I'm a few numbers off, can someone explain to me where I went wrong?
mols glycerin = grams/molar mass = 50/92 = about 0.54 but you can clean that up (as well as the numbers following also).
m = mols/kg solvent
m = 0.54/0.200 = about 2.7
dT = Kf*m = 2.00*2.7 = about 5.4.
New freezing point = -114.6-5.4 = about 120.
To find the freezing point of the solution, you correctly used the equation ∆T = Kf * m, where Kf is the freezing point depression constant and m is the molality of the solution.
First, let's calculate the molality (m) of the solution.
Molar mass of glycerin (C3H8O3) = 92.09 g/mol
Number of moles of glycerin (C3H8O3) = 50.0 g / 92.09 g/mol = 0.543 mol
The molar mass of ethanol (C2H5OH) is 46.07 g/mol.
Number of moles of ethanol (C2H5OH) = 200.0 g / 46.07 g/mol = 4.34 mol
Now, we need to find the total moles of solute in the solution.
Total moles of solute = moles of glycerin + moles of ethanol = 0.543 mol + 4.34 mol = 4.883 mol
Now, we can calculate the molality (m) of the solution.
m = moles of solute / mass of solvent (in kg)
mass of solvent (in kg) = 200.0 g ethanol / 1000 = 0.200 kg
m = 4.883 mol / 0.200 kg = 24.415 mol/kg
Finally, we can find the freezing point depression (∆T) by plugging the values into the equation.
∆T = Kf * m
∆T = 2.00 C/m * 24.415 mol/kg = 48.83 C
To find the freezing point of the solution, subtract the freezing point depression from the freezing point of pure ethanol.
Freezing point of ethanol (-114.6 C) - ∆T (48.83 C) = -163.43 C
However, it is important to note that the freezing point depression constant (Kf) for ethanol is typically given as 1.99 C/molal or 1.99 C/m. Using this value would yield a slightly different result.
∆T = Kf * m = 1.99 C/molal * 24.415 mol/kg = 48.58385 C
Freezing point of ethanol (-114.6 C) - ∆T (48.58385 C) = -163.18385 C
Therefore, the freezing point of the solution prepared by dissolving 50.0 g of glycerin in 200.0 g ethanol is approximately -163.18 C.
To find the freezing point of the solution, you correctly used the formula:
Change in temperature (ΔT) = Kf * m
Where Kf is the molal freezing point depression constant and m is the molality of the solution.
You correctly identified Kf as 2.00 C/m. Now let's calculate the molality (m) of the solution to find the change in temperature:
The formula for molality (m) is:
m = moles solute / kg solvent
First, we need to calculate the moles of glycerin (C3H8O3) in the solution.
Given:
Mass of glycerin (C3H8O3) = 50.0 g
Molar mass of glycerin (C3H8O3) = (3 * 12.01) + (8 * 1.01) + (3 * 16.00) = 92.09 g/mol
Moles of glycerin = mass of glycerin / molar mass of glycerin
Moles of glycerin = 50.0 g / 92.09 g/mol
Now, calculate the moles of ethanol:
Given:
Mass of ethanol = 200.0 g
Molar mass of ethanol = 46.07 g/mol
Moles of ethanol = mass of ethanol / molar mass of ethanol
Moles of ethanol = 200.0 g / 46.07 g/mol
Since glycerin and ethanol are both solutes, their moles are added together to find the total moles of solute:
Total moles of solute = moles of glycerin + moles of ethanol
Now that we have the total moles of solute, we can calculate the molality (m):
m = total moles of solute / kg solvent
Given:
Mass of solvent (ethanol) = 200.0 g = 0.2 kg
Now, divide the total moles of solute by the kilograms of solvent to find the molality (m):
m = (total moles of solute) / (mass of solvent in kg)
Finally, substitute the value for m and Kf into the equation to calculate ΔT:
ΔT = Kf * m
After calculating ΔT, subtract it from the freezing point of pure ethanol (-114.6°C) to find the freezing point of the solution.
Based on the given information, your calculation for molality (m) seems to be incorrect. Please recheck your calculation and make sure you're using the correct masses and molar masses for both the glycerin and ethanol.