delta T = Kb*m
Substitute and solve for m
Then plug m and Kf into your delta T = Kf*m formula and solve for delta T, then freezing point.
I know that formula for freezing point is delta Tf = Kf*m but what do I plug in for each?
Substitute and solve for m
Then plug m and Kf into your delta T = Kf*m formula and solve for delta T, then freezing point.
Kb = 0.512
Kf = 1.86
For boiling point data:
delta T =Kb*m
(102.48-100) = 0.512*m
m = about 4.84m
Then delta T = Kf*m
dT = 1.86*4.84
dT = about9.00 = about 9.00 degrees lower than the normal freezing point.
0-9.00 = -9.00 C.
2.1/0.512=4.10
4.10*1.86=7.62
In the formula, delta Tf represents the change in freezing point, Kf is the freezing point depression constant (which is specific to the solvent), and m is the molality of the solution (the concentration expressed in moles per kilogram of solvent).
Since you're dealing with an aqueous solution, water is your solvent. Now, you just need to find the Kf value for water, and the molality of the solution. Once you have those, you can plug them into the formula and calculate the freezing point!
- delta Tf represents the change in freezing point and is measured in degrees Celsius.
- Kf is the molal freezing point depression constant, which is specific to the solvent being used. For water, this constant is approximately 1.86 °C/m.
- m represents the molality of the solution, which is the number of moles of solute per kilogram of solvent.
In order to plug in the correct values for delta Tf and Kf, we need to know the molality of the solution. Since you haven't provided the molality, we'll assume a hypothetical value.
Let's say the molality of the solution is 0.5 mol/kg. Now we can calculate delta Tf as follows:
delta Tf = Kf * m
delta Tf = (1.86 °C/m) * (0.5 mol/kg)
delta Tf = 0.93 °C
Therefore, the freezing point of the aqueous solution is 0.93 °C below the freezing point of pure water (0 °C). To find the actual freezing point, you would subtract 0.93 °C from 0 °C, resulting in a freezing point of -0.93 °C.