Question

What is the percent by mass of glycerol C3H8O3 in a 2.50 m aqueous solution of glycerol?

I know how to get percent by mass but does the 2.50 m have any relevance to the question?

Answer 1

The 2.50M (I guess is a capital M) means the concentration, Molarity or number of moles of glycerol that you have per liter of solution. You supposed to be given the density of that solution. Otherwise consider density of the water 1g/ml.
Then 2.50M means that you have 2.5moles og glycerol in a liter of solution. You calculate the molecular mass of the glycerol, multiply by 2.5 and that is going to be the mass of glycerol per liter of solution. From that point you know the rest.

Answer 2

It's a lower case italicized 'm' so i don't know if that is the same meaning as the capital 'M.' But once i multiply the molecular mass by 2.5, i just divide each element by what i got from multiplying the molecular mass by 2.5 and then multiply it by 100?

Answer 3

2.50 mg/ml?

Answer 4

no, that will be the mass of glycerol in 1000, just divide by 10 and you will get the mass per 100

Answer 5

Wouldn't lowercase m be molality? so moles/kg?

Answer 6

yeah, \(m\) is molality

Answer 7

Hey @aaronq. So how exactly would you go about this problem?

Answer 8

Find the masses of each

\(molality=\dfrac{n_{solute}}{kg_{solvent}}\)

Assuming you have 1L, you'll have 2.5 moles of glycerol

Convert that to mass and it'd be safe to assume that the density of water is 1 g/mL

Answer 9

i'll give that a try right now

Answer 10

hm i see another problem. you're not told the density of the solution... you'll have to make the assumption that the solute doesn't change the volume of the solution... which might be unrealistic.

Answer 11

would that assumption change how i target this problem?

Answer 12

no not really, it'll just be an estimate. I gotta go, take care.

Answer 13

Quick questions @aaronq !

Answer 14

question* just one

Answer 15

what exactly am i converting to mass? I got the masses for each and the molar mass, but now i'm a bit stuck on what to do next

Answer 16

the molecular mass of the glycerol is 92.09382 g/mol

If we agree that the m is equal to molality as @Jaipal21 @aaronq mention and not Molarity as I was thinking before, that means you will have

(92.09382 g/mol ) * 2.5 mol/kg of solvent = 230.23455 g of glycerol by kg of solvent, so you will have 230.235 g of glycerol in 1230.235 g of solution.

The change in volume is not important because molalidad is defined in function of the mass of solute and mass of solvent and not in function of the volume. And % by mass is also defined in function of mass and not in function of volume. Then you have to add the mass of solute and solvent to have the total mass of solution that is going to be your 100% of mass and the mass of glycerol will be the % by mass that is asking the problem

mass of solution = mass of solute + mass of solvent

Then 230.235g glycerol *100 /1230.235 g of solution = 18.71%

Answer 17

Thank you @cuanchi