How much ethyl alcohol (C2H5OH) must be added to 1.00 L of water so that the solution will not freeze at -4°F?
To determine the amount of ethyl alcohol that must be added to 1.00 L of water so that the solution will not freeze at -4°F, we need to use the concept of freezing point depression.
The freezing point depression is given by the formula:
ΔT = K_f * m
Where:
ΔT = change in freezing point (in degrees Celsius or Kelvin)
K_f = cryoscopic constant (in degrees Celsius/mol or Kelvin/mol)
m = molality of the solute (in mol solute/kg solvent)
First, convert -4°F to Celsius:
-4°F = -20°C
Next, we need to find the molality of the solute. Molality is defined as the number of moles of solute per kilogram of solvent.
Since we're adding ethyl alcohol (C2H5OH), we need to calculate the number of moles of ethyl alcohol. The molar mass of C2H5OH can be calculated as follows:
C: 12.01 g/mol
H: 1.01 g/mol (3 hydrogen atoms)
O: 16.00 g/mol
Total molar mass of C2H5OH = (2*12.01) + (6*1.01) + 16.00 = 46.07 g/mol
To find the number of moles of ethyl alcohol required, we need to know the density of ethyl alcohol. Let's assume it is 0.789 g/mL.
Given that we want to prepare the solution in 1.00 L of water, we can calculate the mass of ethyl alcohol needed:
Mass of ethyl alcohol = desired concentration * volume of solvent
Mass of ethyl alcohol = (desired concentration in g/mL) * (volume of solvent in mL)
The desired concentration can be determined by rearranging the molality formula:
m = (moles of solute) / (mass of solvent in kg)
moles of solute = m * (mass of solvent in kg)
Since we know the mass of the solvent in grams, we need to convert it to kilograms:
Volume of solvent = 1.00 L = 1000 mL
Moles of solute = m * (mass of solvent in kg)
Moles of solute = m * (mass of solvent in grams / 1000)
Using the density of ethyl alcohol, we can find the volume needed:
Volume of ethyl alcohol = (mass of ethyl alcohol) / (density of ethyl alcohol)
Now, let's plug in the values and calculate the quantities.
Given:
Freezing point depression (ΔT) = -20°C
Density of ethyl alcohol = 0.789 g/mL
Volume of solvent = 1000 mL = 1.00 L
Desired concentration = m (to be determined)
Solution:
1. Calculate the molality (m):
m = (moles of solute) / (mass of solvent in kg)
m = ΔT / K_f
Since we don't have the cryoscopic constant (K_f) for water, let's use the average value of 1.86 °C/m.
m = -20°C / 1.86 °C/m
m ≈ -10.8 m
2. Calculate the mass of ethyl alcohol needed:
Mass of solvent = 1000 g
Mass of ethyl alcohol = m * (mass of solvent in grams / 1000)
Mass of ethyl alcohol = -10.8 * (1000 g / 1000)
Mass of ethyl alcohol ≈ -10.8 g
Since a negative mass is not meaningful, it indicates that a solution of ethyl alcohol and water will not freeze at -4°F. In other words, any amount of ethyl alcohol above 10.8 grams will be sufficient to prevent the solution from freezing at -4°F.
To determine the amount of ethyl alcohol (C2H5OH) required to prevent freezing of water at a given temperature, we need to calculate the freezing point depression constant (Kf) for the solvent, in this case, water.
The freezing point depression formula is given as:
ΔT = Kf × m
Where:
ΔT is the change in freezing point (in °C or K)
Kf is the freezing point depression constant (in °C/m or K/m)
m is the molality of the solute (in mol/kg)
First, let's convert -4°F to °C:
°C = (°F - 32) × 5/9
°C = (-4 - 32) × 5/9
°C = -20 × 5/9
°C = -11.1°C
Since we want the solution to not freeze at -11.1°C, this is the value we will use for ΔT.
The freezing point depression constant for water is 1.86 °C/m.
Now, let's calculate the molality (m) of the solution:
m = moles of solute / kg of solvent
For 1.00 L of water, we can assume the density of water is approximately 1.00 kg/L because the density of water is close to 1 g/mL.
Therefore, the molality (m) is:
m = moles of solute / 1.00 kg
Now, let's rearrange the freezing point depression formula to solve for moles of solute:
moles of solute = ΔT / Kf × m
Substituting the known values:
moles of solute = -11.1 / 1.86 × m
To determine the amount of ethyl alcohol in moles, we need to know its molecular weight. The molecular weight of C2H5OH is:
(2 × 12.01 g/mol) + (6 × 1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
Now, let's convert the desired volume of ethyl alcohol into kilograms:
1.00 L = 1000 g (assuming the density of ethyl alcohol is approximately 1 g/mL)
1.00 kg = 1000 g
Now, we can calculate the moles of ethyl alcohol required:
moles of ethyl alcohol = (mass of ethyl alcohol in g) / (molecular weight of ethyl alcohol in g/mol)
moles of ethyl alcohol = (1000 g) / (46.07 g/mol)
Finally, we can substitute this value into the previous equation to find the molality (m) of the solution:
m = moles of ethyl alcohol / 1.00 kg
Now you have the molality of the solution, which can be used to determine the amount of ethyl alcohol (C2H5OH) that must be added to 1.00 L of water to prevent freezing at -4°F.
2.47
delta T = Kf*m
calculate m
m = moles/1 kg solvent
calculate moles
moles = grams/molar mass
calculate grams.
Use the density if you want to measure the ethanol by volume.