. What is the freezing point of a solution that contains 3.5 mol of the ionic solid BaCl2 dissolved
in 500. grams water?
depression=-.52*3.5*2
the molality concentration is 3*3.5*2
To find the freezing point of a solution, we need to use the concept of freezing point depression. The freezing point depression (∆Tf) is the difference between the freezing point of the pure solvent and the freezing point of the solution.
To calculate the freezing point depression, we need to know the molality (moles of solute per kilogram of solvent) and the cryoscopic constant (which is specific to the solvent being used). The cryoscopic constant for water is 1.86 °C·kg/mol.
First, we need to calculate the molality of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Given:
Moles of solute (BaCl2) = 3.5 mol
Mass of solvent (water) = 500. grams = 0.5 kg
Molality (m) = 3.5 mol / 0.5 kg = 7 mol/kg
Next, we can use the molality and the cryoscopic constant to calculate the freezing point depression (∆Tf):
∆Tf = cryoscopic constant (Kf) * molality
For water, the cryoscopic constant (Kf) is 1.86 °C·kg/mol.
∆Tf = 1.86 °C·kg/mol * 7 mol/kg = 13.02 °C
Finally, we can calculate the freezing point of the solution by subtracting the freezing point depression from the freezing point of the pure solvent, which is 0 °C for water.
Freezing point = 0 °C - 13.02 °C = -13.02 °C
Therefore, the freezing point of the solution containing 3.5 mol of BaCl2 dissolved in 500 grams of water is approximately -13.02 °C.