A solution is prepared by placing 72.8 g of calcium chloride in 120 g of water at 22C. The vapor pressure of water at 22C is 2.6 kPa.
a)What is the boiling point of the solution?
b)What is the freezing point of the solution?
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To calculate the boiling point and freezing point of the solution, we need to use the concept of boiling point elevation and freezing point depression. These phenomena occur when a solute is added to a solvent, and the boiling point and freezing point of the solution differ from those of the pure solvent.
a) Boiling Point Elevation:
To find the boiling point elevation of the solution, we can use the equation:
ΔTb = Kb * m
Where:
ΔTb = the change in boiling point
Kb = the molal boiling point elevation constant for the solvent (water)
m = molality of the solution.
First, we need to calculate the molality of the solution:
Moles of calcium chloride = mass / molar mass
Moles of calcium chloride = 72.8 g / (40.08 g/mol + 2 * 35.45 g/mol) = 0.5 mol
Moles of water = mass / molar mass
Moles of water = 120 g / 18.02 g/mol = 6.655 mol
Molality = moles of solute / mass of solvent (in kg)
Molality = 0.5 mol / 0.120 kg = 4.17 mol/kg
Now we can use the equation:
ΔTb = Kb * m
The molal boiling point elevation constant (Kb) for water is 0.512 °C/m.
ΔTb = 0.512 °C/m * 4.17 mol/kg
ΔTb = 2.13 °C
The boiling point elevation is 2.13 °C.
To find the boiling point of the solution, we add the boiling point elevation to the boiling point of the pure solvent (water).
Boiling point of solution = boiling point of pure solvent + boiling point elevation
Boiling point of solution = 100 °C (boiling point of water) + 2.13 °C = 102.13 °C
Therefore, the boiling point of the solution is 102.13 °C.
b) Freezing Point Depression:
To find the freezing point depression of the solution, we can use the equation:
ΔTf = Kf * m
Where:
ΔTf = the change in freezing point
Kf = the molal freezing point depression constant for the solvent (water)
First, we need to calculate the molality of the solution (already calculated above):
Molality = 4.17 mol/kg
The molal freezing point depression constant (Kf) for water is 1.86 °C/m.
ΔTf = 1.86 °C/m * 4.17 mol/kg
ΔTf = 7.755 °C
The freezing point depression is 7.755 °C.
To find the freezing point of the solution, we subtract the freezing point depression from the freezing point of the pure solvent (water).
Freezing point of solution = freezing point of pure solvent - freezing point depression
Freezing point of solution = 0 °C (freezing point of water) - 7.755 °C = -7.755 °C
Therefore, the freezing point of the solution is -7.755 °C.
A solution is prepared by placing 72.8 g of calcium chloride in 120 g of water at 22C. The vapor pressure of water at 22C is 2.6 kPa.
a)What is the boiling point of the solution?
b)What is the freezing point of the solution?