What is the expected freezing point of a 3.50 m (m standing for molality)aqueous solution of BaCl2(Kfp= 1.86 degrees Celsius/m)? What is the expected boiling point(Kbp= 0.512 degrees Celsius/m)? which is effected to the greater extent, the boiling point or freezing point?
Please help. i'm so confused. i don't even know how to start this problem.
6.51
To determine the expected freezing point and boiling point of a solution, you will need to use the concept of colligative properties. Colligative properties depend on the number of solute particles, not their identity. The two most common colligative properties are the freezing point depression and the boiling point elevation.
To start, we need to convert the given molality (m) of the solution to moles of solute per kilogram of solvent by using the formula:
molality (m) = moles of solute / mass of solvent in kg
Given:
Molality (m) = 3.50 m (molality)
Kfp value for water = 1.86°C/m (freezing point constant)
Kbp value for water = 0.512°C/m (boiling point constant)
Let's assume we have 1 kg of water as the solvent.
1. Freezing point depression:
The freezing point depression can be calculated using the formula:
∆Tf = Kfp * m
∆Tf = 1.86°C/m * 3.50 m
∆Tf = 6.51°C
The negative sign indicates that the freezing point will decrease from the normal freezing point of water.
Therefore, the expected freezing point of the solution is the normal freezing point of water (0°C) minus the freezing point depression (-6.51°C):
Expected freezing point = 0°C - 6.51°C = -6.51°C
2. Boiling point elevation:
The boiling point elevation can be calculated using the formula:
∆Tb = Kbp * m
∆Tb = 0.512°C/m * 3.50 m
∆Tb = 1.79°C
The positive sign indicates that the boiling point will increase from the normal boiling point of water.
Therefore, the expected boiling point of the solution is the normal boiling point of water (100°C) plus the boiling point elevation (1.79°C):
Expected boiling point = 100°C + 1.79°C = 101.79°C
Finally, comparing the two values, we can see that the magnitude of the boiling point elevation (1.79°C) is greater than the freezing point depression (-6.51°C). Therefore, the boiling point is affected to a greater extent compared to the freezing point in this solution.
Please note that these calculations assume ideal behavior of the solute and solvent, and any non-ideal interactions may slightly affect the results.