in 0.3m solution of it we have 0.3 moles * molar mass (340g/mol)= 102 grams of Al2(SO4)3 but remember we also have 1000 g of solvent (molality is moles of solute per kilogram of solvent). So total mass is 102+1000 = 1102 grams
So if we have 102 grams of Al2(SO4)3 and the total mass is 1102 then 102/1102 = the percent of Al2(SO4)3 in the whole thing = 0.0925
Now if we have 85.7 grams of total solution how much Al2(SO4)3 is in there? 0.0925 * 87.5
I think you are getting confused. The answer (87.5 x 102.6)/1102.6 is not an equation of molality. The answer shows the proportion of Al2(SO4)3 out of the total mass. But yes molality is mole solute / kg solvent.
For example if you have 10 g A and 15 g B what is the percent of A in the total solution? (10/(10+15)) = 10/25
So if I say a ball of mass of A+B weighs 200g what part of it is A? 200*(10/25)
This is the same as what the problem is asking. Final the total mass. Divide the mass of Al2(SO4)3 by the total mass to find the part of Al2(SO4)3 in it. Then you can use this fraction to find out the percent of Al2(SO4)3 in others.
So above we have 0.3m or 0.3 moles of solute per 1 kilogram of solvent. So the total mass is mass of Al2(SO4)3 from 0.3 moles + 1000 g of solvent.
how do you know that there is 1000g of the solvent? I didn't catch how to determine how much solvent there is in the question? Obviously 1000g = 1kg, but can you just assume that there's going to be 1kg of solvent? I'm a little confused...
how do you know that there is 1000g of the solvent? I didn't catch how to determine how much solvent there is in the question? Obviously 1000g = 1kg, but can you just assume that there's going to be 1kg of solvent? I'm a little confused...
