What is the boiling point and freezing point of a 6.34 m solution of sodium chloride (NaCl) in acetone (kb = 1.71 0C/m; kf = 2.40 0C/m; normal boiling point = 55.9 0C; normal freezing point = -94.8 0C).
Boiling point = 30.6 0C, Freezing point = -77.4 0C
Boiling point = 35.6 0C, Freezing point = -23.4 0C
Boiling point = 46.6 0C, Freezing point = -56.4 0C
Boiling point = 77.6 0C, Freezing point = -125.2 0C
Boiling point = 100.6 0C, Freezing point = -104.4 0C
The boiling point of the solution can be calculated using the formula:
ΔT = kb * m
Where ΔT is the change in boiling point, kb is the molal boiling point elevation constant, and m is the molality of the solution.
Substituting the given values, we have:
ΔT = (1.71 0C/m) * (6.34 mol/kg) = 10.8374 0C
The boiling point of the solution is the normal boiling point (55.9 0C) minus the change in boiling point (10.8374 0C):
Boiling point = 55.9 0C - 10.8374 0C = 45.0626 0C ≈ 45.1 0C
Similarly, the freezing point of the solution can be calculated using the formula:
ΔT = kf * m
Where ΔT is the change in freezing point, kf is the molal freezing point depression constant, and m is the molality of the solution.
Substituting the given values, we have:
ΔT = (2.40 0C/m) * (6.34 mol/kg) = 15.216 0C
The freezing point of the solution is the normal freezing point (-94.8 0C) plus the change in freezing point (15.216 0C):
Freezing point = -94.8 0C + 15.216 0C = -79.584 0C ≈ -79.6 0C
Therefore, the correct answer is:
Boiling point = 45.1 0C
Freezing point = -79.6 0C