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.
C=5/9(F-32)=-160.9=-17.7 C
-17.7C=-1.86 (molesethantol/.5kg)
solve for moles ethanol, convert to grams.
I get about 4.75moles, or about 220 grams.
To answer this question, we need to use the formula for freezing point depression:
ΔT = K_f * m
where:
ΔT is the change in freezing point (in degrees Celsius or Kelvin)
K_f is the molal freezing point depression constant
m is the molality of the solute (moles of solute per kilogram of solvent)
First, we need to convert the freezing point from Fahrenheit to Celsius since the freezing point depression constant is given in degrees Celsius. To convert Fahrenheit to Celsius, we use the formula:
°C = (°F - 32) / 1.8
Given that the freezing point is 0.00 degrees Fahrenheit, we convert it to Celsius:
°C = (0.00 - 32) / 1.8
°C = -17.78
Next, we need to calculate the molal concentration (molality) of the solute. Molality is defined as the moles of solute (ethanol) per kilogram of solvent (water).
We know the mass of water is 500.0 g, which can be converted to kilograms:
mass of water = 500.0 g = 0.500 kg
Now, we need to calculate the moles of ethanol required to achieve the desired freezing point depression.
ΔT = K_f * m
Since the desired ΔT is -17.78°C (-17.78 K) and K_f is given as 1.86°C/mol:
-17.78 = 1.86 * m
Solving for m:
m = -17.78 / 1.86
m ≈ -9.57 mol/kg
Since molality cannot be negative, we take the absolute value:
m ≈ 9.57 mol/kg
Now that we know the molality of ethanol, we can calculate the moles of ethanol needed using the formula:
moles of ethanol = molality * mass of water (in kg)
moles of ethanol = 9.57 mol/kg * 0.500 kg
moles of ethanol ≈ 4.79 mol
Finally, we need to convert moles of ethanol to grams:
mass of ethanol = moles of ethanol * molar mass of ethanol
The molar mass of ethanol (C2H6O) is approximately 46.07 g/mol, so:
mass of ethanol = 4.79 mol * 46.07 g/mol
mass of ethanol ≈ 219.85 g
Therefore, approximately 219.85 grams of ethanol must be added to 500.0 g of water to make a solution that freezes at 0.00 degrees Fahrenheit.