What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 20.0 L of water to produce an antifreeze solution with a freezing point of -28.0°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)
.....L
What is the boiling point of this solution?
......°C
-28=-.52molesglycol/kg solution
kg solution=massglycol+masswater
-28=-0.52(massglycol/molmassglycol)/(massglycol/1000 + 20kg)
where massglycol is in grams.
To determine the volume of ethylene glycol that must be added to water to produce an antifreeze solution with a specific freezing point, we can use the concept of molality.
The molality of a solution is defined as the moles of solute per kilogram of solvent. In this case, the solute is ethylene glycol (C2H6O2) and the solvent is water.
First, we need to convert the given volume of water from liters to kilograms. The density of water is 1.00 g/cm3, so the mass of water can be calculated as follows:
Mass of water = density of water * volume of water
Mass of water = 1.00 g/cm3 * 20.0 L * (1000 cm3/1 L) * (1 kg/1000 g)
Mass of water = 20.0 kg
Next, we need to determine the molality of the solution required to achieve a freezing point of -28.0°C. The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = kf * m
where kf is the freezing point depression constant for water (1.86°C/m) and m is the molality of the solution.
In this case, we can rearrange the formula to solve for m:
m = ΔTf / kf
m = (-28.0°C - 0°C) / (-1.86°C/m)
m = 15.05 m
Now, we can calculate the moles of ethylene glycol required to achieve the desired molality. The molecular weight of ethylene glycol (C2H6O2) is 62.07 g/mol.
moles of ethylene glycol = m * (mass of water in kg) / molecular weight of ethylene glycol
moles of ethylene glycol = 15.05 m * 20.0 kg / 62.07 g/mol
moles of ethylene glycol = 4.857 mol
Finally, we can calculate the volume of ethylene glycol using its density. The density of ethylene glycol is 1.11 g/cm3.
Volume of ethylene glycol = (moles of ethylene glycol) * (molecular weight of ethylene glycol) / (density of ethylene glycol)
Volume of ethylene glycol = 4.857 mol * 62.07 g/mol / 1.11 g/cm3
Volume of ethylene glycol = 272.932 cm3 or 0.273 L
Therefore, to produce an antifreeze solution with a freezing point of -28.0°C, approximately 0.273 L of ethylene glycol must be added to 20.0 L of water.
To determine the boiling point of the solution, we can use the formula for the boiling point elevation (ΔTb):
ΔTb = kb * m
where kb is the boiling point elevation constant for water (0.512°C/m) and m is the molality of the solution.
Using the calculated molality (15.05 m):
ΔTb = 0.512°C/m * 15.05 m
ΔTb = 7.7068°C
The boiling point of the solution can be determined by adding the boiling point elevation to the boiling point of pure water (100°C):
Boiling point of solution = 100°C + 7.7068°C
Boiling point of solution ≈ 107.7°C
Therefore, the boiling point of this antifreeze solution is approximately 107.7°C.
To calculate the volume of ethylene glycol required to produce an antifreeze solution with a freezing point of -28.0°C, we can use the equation for freezing point depression:
ΔT = Kf * molality
Where:
ΔT is the change in freezing point temperature
Kf is the freezing point depression constant
molality is the molal concentration of the solute (ethylene glycol)
Given that the freezing point depression constant (Kf) for water is 1.86 °C/m and the freezing point of the solution is -28.0 °C lower than that of pure water, we have:
ΔT = -28.0 °C
To find molality, we need to convert grams of ethylene glycol to moles and calculate the molal concentration:
Molar mass of ethylene glycol (C2H6O2) = (2*12.01 g/mol) + (6*1.01 g/mol) + (2*16.00 g/mol) = 62.07 g/mol
Given that the density of ethylene glycol is 1.11 g/cm3, we can calculate the volume of ethylene glycol required:
Volume of ethylene glycol = mass of ethylene glycol / density of ethylene glycol
mass of ethylene glycol = volume of ethylene glycol * density of ethylene glycol
Substituting the values:
mass of ethylene glycol = (volume of water in liters * density of water in g/cm3) * (1 cm3 / 1 mL) * (1 mL / 1 cm3) * (1000 g / 1 kg)
mass of ethylene glycol = (20.0 L * 1000 g/L) * (1 cm3 / 1 mL) * (1 mL / 1 cm3) * (1 kg / 1000 g) = 20000 g
Now we can calculate the molality:
molality = moles of solute / mass of solvent (in kg)
moles of solute = mass of ethylene glycol / molar mass of ethylene glycol
moles of solute = 20000 g / 62.07 g/mol = 322.2 mol
mass of solvent = volume of water in liters * density of water in g/cm3 * (1 cm3 / 1 mL) * (1 mL / 1 cm3) * (1 kg / 1000 g)
mass of solvent = (20.0 L * 1000 g/L) * (1 cm3 / 1 mL) * (1 mL / 1 cm3) * (1 kg / 1000 g) = 20000 g
molality = 322.2 mol / 20.0 kg = 16.1 mol/kg
Now, we can substitute the values into the freezing point depression equation:
ΔT = Kf * molality
-28.0 °C = 1.86 °C/m * 16.1 mol/kg
Solving for Kf:
Kf = -28.0 °C / (1.86 °C/m * 16.1 mol/kg) = -0.277 °C/m
The boiling point elevation equation is similar to the freezing point depression equation:
ΔT = Kb * molality
where:
ΔT is the change in boiling point temperature
Kb is the boiling point elevation constant
molality is the molal concentration of the solute (ethylene glycol)
For water, the Kb value is typically 0.512 °C/m.
Using this value, we can calculate the boiling point elevation:
ΔT = Kb * molality
ΔT = 0.512 °C/m * 16.1 mol/kg = 8.2612 °C
Now, we can calculate the boiling point of the solution:
Boiling point of the solution = boiling point of pure solvent (water) + ΔT
Boiling point of pure water at standard atmospheric pressure is 100.0 °C.
Boiling point of the solution = 100.0 °C + 8.2612 °C = 108.26 °C
Therefore, the boiling point of the solution is approximately 108.26 °C.