a) An ethylene glycol solution contains 21.1 g of ethylene glycol in 86.0 ml of water. compute the freezing point of the solution.
b) Compute the boiling point of the solution.
moles ethylene glycol = grams/molar mass.
molality = moles/kg solvent.
Solve for molality.
delta T = Kf*m
Find freezing point form delta T knowing that the normal freezing point of water is zero C.
Same procedure for boiling point except
delta T = Kb*m
and add delta T to 100 C to find boiling point.
what is Kf and Kb
Kf is the freezing point depression constant and it should be in your text or notes. Kb is the boiling point elevation constant and it should be available to you also. Kf in my text = 1.86 for water and Kb is 0.512 for H2O.
I'm terrible at these!!! I wish I could help you.
To compute the freezing point of the solution, we need to use the concept of freezing point depression. The formula for calculating freezing point depression is given by:
ΔT = K_f * m
Where:
ΔT is the change in freezing point (in degrees Celsius),
K_f is the cryoscopic constant (specific to a given solvent),
m is the molality of the solution, which is the moles of solute per kilogram of the solvent.
a) Finding the molality of the solution:
First, we need to determine the moles of ethylene glycol (solute) and water (solvent) separately.
Molar mass of ethylene glycol (C2H6O2):
C: 12.01 g/mol x 2 = 24.02 g/mol
H: 1.01 g/mol x 6 = 6.06 g/mol
O: 16.00 g/mol x 2 = 32.00 g/mol
Total molar mass = 24.02 g/mol + 6.06 g/mol + 32.00 g/mol = 62.08 g/mol
Moles of ethylene glycol = mass / molar mass
Moles of ethylene glycol = 21.1 g / 62.08 g/mol = 0.3395 mol
Now, let's calculate the molality of the solution using water as the solvent:
Mass of water = volume of water x density of water
Volume of water = 86.0 ml = 0.086 L (since 1 mL = 0.001 L)
Density of water = 1 g/mL (approximately)
Mass of water = 0.086 L x 1 g/mL = 0.086 kg
Molality (m) = moles of ethylene glycol / mass of water (in kg)
Molality (m) = 0.3395 mol / 0.086 kg = 3.95 m
Now, we need to find the cryoscopic constant for water (K_f). The cryoscopic constant for a solvent can be found in reference tables and is generally given in degrees Celsius per molality (°C/m). For water, the value is approximately 1.86 °C/m.
Substituting the values into the freezing point depression formula:
ΔT = K_f * m
ΔT = 1.86 °C/m * 3.95 m
ΔT = 7.347 °C (rounded to three decimal places)
The freezing point of the solution can be determined by subtracting the freezing point depression from the freezing point of pure water, which is 0 °C.
Freezing point of the solution = 0 °C - 7.347 °C = -7.347 °C
Therefore, the freezing point of the ethylene glycol solution is approximately -7.347 °C.
b) To compute the boiling point of the solution, we use the concept of boiling point elevation. The formula for calculating boiling point elevation is similar to the freezing point depression formula:
ΔT = K_b * m
Where:
ΔT is the change in boiling point (in degrees Celsius),
K_b is the ebullioscopic constant (specific to the solvent),
m is the molality of the solution.
The ebullioscopic constant for water (K_b) is approximately 0.512 °C/m.
Substituting the values into the boiling point elevation formula:
ΔT = K_b * m
ΔT = 0.512 °C/m * 3.95 m
ΔT = 2.022 °C (rounded to three decimal places)
The boiling point of the solution can be determined by adding the boiling point elevation to the boiling point of pure water, which is 100 °C.
Boiling point of the solution = 100 °C + 2.022 °C = 102.022 °C
Therefore, the boiling point of the ethylene glycol solution is approximately 102.022 °C.