What is the freezing point of an aqueous 2.65 m calcium chloride (CaCl2) solution? The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m.
You have three ions in the solution (one Ca++, two Cl-).
freezingpoint= -1.86*3*2.65
To determine the freezing point of an aqueous solution, you can use the equation:
ΔT = Kf * m
Where:
ΔT is the change in temperature (in this case, freezing point depression)
Kf is the cryoscopic constant (given as -1.86ºC/m for pure water)
m is the molality of the solution (in moles of solute per kilogram of solvent)
First, calculate the molality (m) of the solution. In a 2.65 m calcium chloride (CaCl2) solution, the molality indicates that there are 2.65 moles of CaCl2 dissolved per kilogram of water.
Next, substitute the values into the equation:
ΔT = (-1.86ºC/m) * (2.65 m)
Calculate the result:
ΔT = -4.921ºC
To find the freezing point of the solution, subtract the change in temperature from the freezing point of pure water:
Freezing point = 0.0ºC - 4.921ºC
Therefore, the freezing point of the 2.65 m calcium chloride solution is approximately -4.92ºC.