00:01
Hi there.
00:02
The equation used to calculate boiling point elevation and freezing point depression, the two equations are very similar.
00:10
And they look like this, starting with the boiling point.
00:13
To calculate the change in boiling point, we need to take the boiling constant, or the boiling constant, which is for the solvent, and is given to us in this problem, times the molality, times the vantahaw factor, which is i.
00:30
The vantahaw factor is how many, particles the solute breaks into.
00:35
If the solute is ionic, it'll break into two or more particles, and i will be two or however many ions, the ionic compound breaks into.
00:43
If the compound is molecular, it does not break apart in water, and so i is one.
00:49
They told us it is a non -volatile solute, so that means it does not break apart in water, so i is just going to be one in this problem.
00:57
The equation used to calculate freezing point change, freezing point elevation, is very similar, the change in the freezing temperature is equal to kf, which is again a constant.
01:09
It will be different than kb, but it is still based upon the solute, times the molality of the solution, times the vantauff factor.
01:19
Okay, so we are going to first need to solve the delta tb equation for molality, and then use that molality in the delta tf equation to find how much the temperature changes.
01:31
All right, so delta tb, we are told the new boiling point is 102 .5 degrees celsius.
01:43
The solvent is water.
01:45
We're told it's an aquaous solution.
01:47
Water typically freezes, its normal freezing point, is 100 .0 degrees celsius.
01:54
So the difference between this will be the change in the boiling point.
01:59
Kb, the boiling constant for water, we are given as 0 .52.
02:07
Degrees celsius per molal, and the molality is what we are looking for.
02:13
I, we already discussed, is only going to be one, so it's not going to change much here in this equation.
02:19
All right, so the change between the new boiling point and the normal boiling point is going to be 2 .5 degrees celsius.
02:27
This is equal to 0 .52 degrees celsius per molal times the molality.
02:37
I'm going to go ahead and divide both sides by the point five two to isolate molality by itself.
02:47
On the left side of the equation the unit degree celsius will cancel.
02:54
On the right side of the equation, the entire term will cancel.
02:58
So that leaves molality by itself.
03:01
And i am not going to round to significant digits at this point...