00:01
To determine the molality from mass percent, we recognize that 25 % ethylene glycol means we have 25 grams ethylene glycol per 100 grams of solution.
00:18
But we want molality.
00:20
Molality requires moles in the numerator and just kilograms of solvent in the denominator.
00:26
Well, if we have 100 grams of solution and 25 grams of it is ethylene glycol, the remaining 75 grams would be the solvent.
00:36
So we have 75 grams solvent or h2o, which we can convert into kilograms, knowing that that one kilogram is 1000 grams, and we can convert the grams ethylene glycol into moles ethylene glycol using the molar mass of ethylene glycol.
01:07
The molar mass of ethylene glycol is 62 .07 grams for every one mole.
01:19
This then gives us moles ethylene glycol in the numerator and kilograms of just water, the solvent, in the denominator.
01:28
25 divided by 62 .07 divided by what's in the denominator gives us 5 .37 molal or moles ethylene glycol per kilogram h2o.
01:54
Now that we know the molality, we can calculate the boiling point of the solution...