00:01
Question asks you to calculate the mass fraction, mole fraction, molality, and freezing point of the 1 .37 molar equisolution of citric acid.
00:14
Now we were given all different data just to calculate this.
00:17
So first question you want to calculate the mass fraction.
00:23
So the density we call roll of the solution is 1 .10 gram per mil.
00:30
And we know v is 1 liter or 1 ,000 militer.
00:34
So the mass of the entire solution is density multiplied by volume.
00:59
So now you want to calculate the mass of the citric acid.
01:08
The molality of the solution is m equals 1 .37.
01:13
A molarity basically is how many more in one liter.
01:19
So the mole of citric acid is going to be the molarity, 1 .37 mole per liter multiplied by the volume, 1 ,000 mil, or let's say 1 liter .37 mole.
01:41
So if you want to know the mass of citric acid, you're going to have to take the mole multiplied by its molar mass.
01:53
We know that the molar mass of the citric acid, 192.
02:05
So this is the molar mass mm of citric acid.
02:05
This is about 200 ,000 gram per mole.
02:07
So this is the molar mass, mm, of citric acid.
02:13
This is about 200.
02:14
163 .2 gram.
02:20
Now you have a mass fraction can be calculated by the mass of the citric acid divided by the mass of water.
02:38
Mass of water is the total solution 110 gram minus the amount of citric acid.
02:49
That would give you 836 .8 gram.
02:56
Now once we have all that, we have the mass of citric acid and we have a mass of water.
03:04
The mass fraction is the mass of solute, which is the citric acid.
03:19
And then we have the mass of the solution.
03:25
And we know that's 110 .00.
03:28
So this is 260.
03:31
63 .2 gram divided by 1100 gram.
03:36
Remember this is the solution, not the solvent, and multiply by 100 % if you want to...