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.
What concentration of aqueous CaCl2 solution freezes at -10.2ºC? 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 solution, we can use the equation:
ΔT = Kf * molality
ΔT represents the change in freezing point from the freezing point of pure water.
Kf is the cryoscopic constant, which is -1.86 ºC/m for water.
molality (m) is the amount of solute (in moles) divided by the mass of the solvent (in kg).
First, we need to calculate the molality of the potassium nitrate solution:
Molar mass of KNO3 = atomic mass of potassium (39.1 g/mol) + 1 atomic mass of nitrogen (14.0 g/mol) + 3 atomic masses of oxygen (16.0 g/mol) = 39.1 + 14.0 + (3 * 16.0) = 101.1 g/mol
To convert the molarity to molality, we need to know the density of the solution. Let's assume it is 1.00 g/mL.
Mass of KNO3 = 2.25 mol * 101.1 g/mol = 227.475 g
Mass of water = 1000 g (assuming 1 L of solution, which is equivalent to 1000 mL)
Mass of solvent (water) = mass of water / density = 1000 g / 1.00 g/mL = 1000 g
Molality = mol / kg = 2.25 mol / (1000 g / 1000) = 2.25 mol / 1 kg = 2.25 m
Now, we can calculate the change in freezing point:
ΔT = Kf * molality = -1.86 ºC/m * 2.25 m = -4.185 ºC
Finally, we can find the freezing point of the potassium nitrate solution:
Freezing point = Freezing point of water + ΔT = 0.0 ºC - 4.185 ºC = -4.185 ºC
Therefore, the freezing point of the aqueous 2.25 m potassium nitrate (KNO3) solution is approximately -4.185 ºC.
To calculate the freezing point of a solution, we can use the equation:
ΔT = i * Kf * m
Where:
ΔT is the change in freezing point,
i is the van't Hoff factor (the number of particles the solute dissociates into),
Kf is the cryoscopic constant (given as -1.86ºC/m for water),
and m is the molality of the solution (moles of solute per kilogram of solvent).
In this case, the solute is potassium nitrate (KNO3), so we need to determine the van't Hoff factor and the molality.
The van't Hoff factor for KNO3 is 2, as it dissociates into two ions (K+ and NO3-) when it dissolves in water.
Next, we need to determine the molality of the solution. Molality is calculated by dividing the moles of solute by the mass of the solvent (in kilograms). To find the moles of KNO3, we can use the molarity (2.25 M) and the volume of the solution.
Assuming a 1-liter solution for simplicity, we have:
Volume of solution = 1 liter = 1 kg of water (since 1 liter of water has a mass of approximately 1 kg)
To find the moles of KNO3:
moles of KNO3 = molarity * volume (in liters)
= 2.25 M * 1 L
= 2.25 moles
Since the mass of water is 1 kg, the molality (m) of the solution is:
m = moles KNO3 / mass water (in kg)
= 2.25 moles / 1 kg
= 2.25 mol/kg
Now, we can plug the values into the equation:
ΔT = 2 * (-1.86ºC/m) * (2.25 mol/kg)
Simplifying the equation:
ΔT = -2 * 1.86ºC * 2.25 mol/kg
ΔT = -8.37ºC
Finally, to find the freezing point, we subtract the change in freezing point from the freezing point of pure water (0.0ºC):
Freezing point = 0.0ºC - (-8.37ºC)
Freezing point = 8.37ºC
Therefore, the freezing point of the 2.25 m potassium nitrate (KNO3) solution is 8.37ºC.