00:01
We are asked to find the molality, the molarity, and mole fraction.
00:17
Hope i don't regret doing this for 40 % ethylene glycol, which is c2h6o2.
00:41
This is by mass, and i'm going to abbreviate this eg.
00:50
The density of the solution is equal to 1 .75 grams per cubic centimeter.
00:59
Okay, that should be everything i need.
01:04
Okay, the mass percentage by molar mass equals 62 .07 moles.
01:19
So let's get our mass percent.
01:24
Mass of solute and mass of solvent, mass of solution, i should say, we will do as follows.
01:55
I'm going to take my, from my, this one, my mass of my solvents, i'm going to take 0 .4, that's my percent, times my molar, or times my density, which was 1 .75 grams per centimeter cubed, and we multiply that by 1 ,000 centimeters cubed.
02:22
I'm going to divide that by 62 .07 grams per mole.
02:31
This will be grams go out and centimeters cubed, go up.
02:33
Okay? this will be 0 .4 times 1 .75 times 1 ,000 divided by 62 .07.
02:45
And that will equal 11 .2776 moles of ethylene glycol.
03:00
Now let's do how many moles that is.
03:13
That is how many moles? malarity is equal to moles per liter.
03:36
How many liters of solution do i have? i know how many moles of ethylene glycol i have.
03:58
One liter of solution.
04:06
And this is, i think that is actually moles.
04:13
I think that's right.
04:14
I think that's moles per cubic decimeter or bowls per the volume of my solution.
04:24
And i just say that this is one, i'm going to say 11 .277.
04:30
6 divided by 1 liter, and that'll be 11.
04:39
I can go to 3 sig figs, 3 molar.
04:45
That's my molarity.
04:48
Now, the next thing we have to figure out is molality.
04:55
Okay, molality will be moles divided by kilogram of solvent, which in this case is water.
05:04
So if i have 40 kilograms of ethylene glycol per 100 kilograms of solution, that means i have 60 kilograms of water.
05:39
I feel like that's right.
05:48
But i can try this.
05:52
Let's try this a different way...