I know looking at the density before a solution is made with liquids is important, but what about powders?
Lets say someone needed to make a 10% solution of sodium carbonate or a 6% solution of potassium permanganate would they just dissolve 10 gm or 6 gm
into 100 ml of water?
[Edited on 14-2-2006 by T=HC2]Magpie - 13-2-2006 at 19:38
Let's assume you are talking only wt% here. In this case neither the density of the liquid or the powder matters. And if you add 6g powder
to 100g water you don't get a 6 wt% solution, but:
wt% = [6/(6+100)]100% = 5.66%T=HC2 - 14-2-2006 at 21:32
Oh, ok..thanks. It would be by weight then. Can you direct me to a good instructional website that discuses how to figure out how much should be added
for solutions by volume and weight? I don't want to bother you guys anymore with this..lol.
[Edited on 16-2-2006 by T=HC2]Pommie - 15-2-2006 at 07:40
I thought that I understood this.
I thought:-
if I take a bit of water (50mL) and add 5g of salt.
Add water until I have 100mL.
I have a 5% solution.
Having typed that and thought about it, I now understand the previous posts.
Actually, having thought again, the above solution doesn't contain 95g water - or I think it won't - but a little more than that.
Simple question. If I take 100mL of a 5% solution of salt and boil it dry, will I get 5g of salt?
Mike.DrP - 15-2-2006 at 08:22
It depends on the density of the salt Pommie. If the density is 1.00, then yes, but it is not. (Not sure what the density of salt is)Pommie - 15-2-2006 at 08:33
Quote:
Originally posted by DrP
It depends on the density of the salt Pommie. If the density is 1.00, then yes, but it is not. (Not sure what the density of salt is)
As I said, I thought I understood.
So, is there no answer to the question. 100mL of 5% solution when boiled dry will weigh what? Assuming all dissolved components stay to get weighed.
Logically, if 100mL contains 5g of salt ( any salt) then boiling it dry should produce 5g.
Mike.DrP - 15-2-2006 at 08:53
If the 100 ml contains 5g of salt then yes you should get 5 g of salt left behind. This would have been a 5% solution by weight. i.e. 5g
salt, 95g water.
If the solution was a 5% salt solution by volume then the density of the salt would need to be known. i.e. 5 ml of salt and 5 ml of water.
Therefore, lets say the density of salt is 1.2 for arguments sake, you would get 6 g of salt remaining.Pommie - 15-2-2006 at 09:32
Quote:
If the 100 ml contains 5g of salt then yes you should get 5 g of salt left behind. This would have been a 5% solution by weight. i.e. 5g salt, 95g
water.
But, how can 95mL + 5g of salt = 100mL?
Ok, different question. If I see on a bottle 5% saline solution and it's 100mL. How many grams of salt will I get if I boil it dry? I.E. how is it
normally quoted?
Mike.garage chemist - 15-2-2006 at 09:47
Always by weight (except for ethanol, which is given as volume %).
Adding 5ml of salt to 95g water will make a solution of exactly 5% concentration, although its volume is smaller than 100ml.
This is because the density increases.
But as it weighs 100g and contains 5g of salt, it is 5% concentration.
The WEIGHT is important, the volume doesn't play any role.
If making solutions with something different than water, e.g. liquids with a density different than 1, the liquids must always be weighed or their
weight calculated from volume and density.
If you boil down 100ml of a 5% salt solution, you will get slightly more than 5g, because the density of the solution was slightly larger than 1,
giving you a bit more than 100g of solution to start with.
[Edited on 15-2-2006 by garage chemist]T=HC2 - 15-2-2006 at 22:43
Ok, so I would add 6 g to 94 mL water for a 6% solution or 10 g to 90 mL water for a 10% solution by weight?mick - 18-2-2006 at 15:05
I would stick with the idea that 1 formula wt or molecular wt made up to 1 liter is a 1 molar solution in the solvent that is being used.
