What is the freezing point of an aqueous 2.25 m potassium nitrate (KNO3) solution? The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m.
To find the freezing point of the aqueous 2.25 m potassium nitrate (KNO3) solution, we can use the formula:
ΔT = Kf * m
Where:
ΔT is the change in freezing point
Kf is the cryoscopic constant for water, which is -1.86ºC/m
m is the molality of the potassium nitrate (KNO3) solution
First, let's calculate the molality (m) of the potassium nitrate solution. Molality is defined as the number of moles of solute (in this case, KNO3) divided by the mass of the solvent (in this case, water) in kilograms.
Given that the solution is 2.25 m, we know that there are 2.25 moles of KNO3 in every kilogram of water. This can be expressed as:
m = 2.25 mol KNO3 / 1 kg H2O
The freezing point depression, ΔT, can now be calculated using the formula mentioned above:
ΔT = Kf * m
ΔT = (-1.86ºC/m) * (2.25 mol KNO3 / 1 kg H2O)
Simplifying this equation gives us:
ΔT = -1.86ºC * 2.25
ΔT = -4.185ºC
To find the freezing point of the solution, we subtract the ΔT from the freezing point of pure water:
Freezing point of solution = Freezing point of pure water - ΔT
Freezing point of solution = 0.0ºC - (-4.185ºC)
Freezing point of solution = 4.185ºC
Therefore, the freezing point of the aqueous 2.25 m potassium nitrate solution is 4.185ºC.
To determine the freezing point of the aqueous potassium nitrate solution, we can use the concept of freezing point depression. Freezing point depression is the phenomenon where the presence of solute particles in a solvent lowers its freezing point.
The formula to calculate the freezing point depression is:
ΔTf = Kf * m
Where:
- ΔTf is the change in freezing point
- Kf is the cryoscopic constant
- m is the molality of the solution (moles of solute per kilogram of solvent)
First, we need to calculate the molality (m) of the solution. Molality is calculated by dividing the moles of solute (KNO3) by the mass of the solvent (water in this case) in kilograms.
Next, we can substitute the values into the formula to find the change in freezing point (ΔTf).
Finally, we can determine the freezing point of the solution by subtracting the change in freezing point (ΔTf) from the freezing point of pure water, which is 0.0ºC.
Let's plug in the values and calculate:
Given:
- Kf of pure water = -1.86ºC/m
- molality (m) = 2.25 m (moles of KNO3 per kilogram of water)
Step 1: Calculate molality (m)
To find the moles of KNO3, we need to know the molar mass. The molar mass of KNO3 is:
K (atomic mass) = 39.1 g/mol,
N (atomic mass) = 14.0 g/mol,
O (atomic mass) = 16.0 g/mol.
So, molar mass of KNO3 = (39.1 + 14.0 + (3 * 16.0)) g/mol = 101.1 g/mol.
Next, we convert the mass of KNO3 in the solution to moles:
2.25 m = 2.25 moles KNO3 / 1 kilogram water.
Step 2: Calculate ΔTf (change in freezing point)
ΔTf = Kf * m = -1.86ºC/m * 2.25 m = -4.185ºC.
Step 3: Calculate the freezing point of the solution
Freezing point = 0.0ºC (pure water) - 4.185ºC (ΔTf) = -4.185ºC.
Therefore, the freezing point of the aqueous 2.25 m potassium nitrate solution is approximately -4.185ºC.