How many grams of ethanol must be added to 500.0 g of water to make a solution that freezes at 0.00 degrees fahrenheit? The molal freezing point depression constant for water is 1.86 degree celsius over moles.
Convert 0 F to C.
C = (F-32)*5/9 = ?
That's approx -17.78
delta T = Kf*m
17.78 = 1.86*m
Solve for molality.
Then m = mol/kg solvent
You have m and kg H2O, solve for mols ethanol.
Finally, mols ethanol = grams/molar mass. You have molar mass and mols, solve for grams.
Post your work if you get stuck.
220 g
Well, you're in luck! I happen to know a joke that's related to your question. Why did the scarecrow win an award? Because he was outstanding in his field!
Now, let's get back to your question. We can use the formula for freezing point depression to solve this. The formula is:
∆T = Kf * m
where ∆T is the change in freezing point, Kf is the molal freezing point depression constant for water, and m is the molality of the solute.
First, we need to convert 0.00 degrees Fahrenheit to degrees Celsius since our molal freezing point depression constant is given in Celsius. We know that 1 degree Fahrenheit is equal to 0.556 degrees Celsius, so 0.00 degrees Fahrenheit is approximately -17.78 degrees Celsius.
Now, let's plug in the values. We want the freezing point of the solution to be -17.78 degrees Celsius, so:
∆T = -17.78 - 0 = -17.78 degrees Celsius
Kf = 1.86 degrees Celsius/moles
m = ?
500.0 g of water is approximately equal to 0.556 moles.
Plugging in the values into the formula, we get:
-17.78 = 1.86 * m
Now we can solve for m:
m = -17.78 / 1.86
After doing the math, we find that m is approximately -9.57. Since molality cannot be negative, it seems like we've run into a bit of a problem. It looks like we need imaginary ethanol to make the solution freeze at -17.78 degrees Celsius. Maybe science should invent some imaginary ethanol so we can have imaginary cold drinks!
I hope this humorous explanation lightened up the mood a bit, even though we couldn't find a feasible solution. If you have any more questions or need a laugh, feel free to ask!
To solve this problem, we need to use the formula for freezing point depression:
ΔTf = Kf * molality
Where:
- ΔTf is the freezing point depression (in degrees Celsius)
- Kf is the molal freezing point depression constant for water (1.86 °C/m)
- molality is the moles of solute per kilogram of solvent (mol solute/kg solvent)
First, we need to convert the freezing point to degrees Celsius because the molal freezing point depression constant is given in °C.
0.00 degrees Fahrenheit = -17.78 degrees Celsius
Next, we need to calculate the molality of the solution. Molality is defined as the moles of solute per kilogram of solvent.
Given that the solute is ethanol (C2H5OH) and the solvent is water (H2O), we need to calculate the number of moles of ethanol and the total mass of the solvent (water).
The molar mass of ethanol (C2H5OH) is:
2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
To calculate the moles of ethanol, we divide the mass of ethanol by the molar mass:
moles of ethanol = mass of ethanol / molar mass of ethanol
Next, we determine the total mass of the solution (ethanol + water):
mass of solution = mass of ethanol + mass of water
Now, we can calculate the molality using the formula:
molality = moles of ethanol / mass of water (in kg)
Lastly, we can use the freezing point depression formula to find the mass of ethanol required to achieve a freezing point depression of -17.78 degrees Celsius (0.00 degrees Fahrenheit).
ΔTf = Kf * molality
-17.78 °C = 1.86 °C/m * molality
Now that we have the molality and the freezing point depression, we can determine the grams of ethanol required to achieve the desired freezing point depression.