Answer:
the freezing point of the 1.17 m CaCl₂ solution is -2.18 °C and the boiling point is 100.599 °C.
Explanation: The freezing and boiling points of an aqueous solution can be calculated using the following equations:
ΔTf = Kf * m
ΔTb = Kb * m
where:
ΔTf is the change in freezing point
ΔTb is the change in boiling point
Kf is the freezing point depression constant
Kb is the boiling point elevation constant
m is the molality of the solution
For water, Kf = 1.86 °C/m and Kb = 0.512 °C/m.
For a 1.17 m CaCl₂ solution:
ΔTf = Kf * m
ΔTf = 1.86 °C/m * 1.17 m
ΔTf = 2.18 °C
The freezing point depression is 2.18 °C. The freezing point of pure water is 0 °C, so the freezing point of the CaCl₂ solution is:
Freezing point = 0 °C - 2.18 °C
Freezing point = -2.18 °C
ΔTb = Kb * m
ΔTb = 0.512 °C/m * 1.17 m
ΔTb = 0.599 °C
The boiling point elevation is 0.599 °C. The boiling point of pure water is 100 °C, so the boiling point of the CaCl₂ solution is:
Boiling point = 100 °C + 0.599 °C
Boiling point = 100.599 °C