[Traumatic Acid]

Hi all
This could be a stupid question.
If I had a solution made up of two parts, water and an unknown concentration of an unknown solute, can I use this method to calculate the % by weight of the solute?

If all that is known about the solution is that the solvent is pure water:
Can I weigh 1000 ml of the solution (say it weighs 1360 grams) then deduce that the solution contains 360 grams per litre solute(s)?
I know that I could just weigh one ml, and I know that water does not have a density of precisely one. But in theory this method should work right? How about in practice?

Cheers

[Borek]

36% solution of AgNO₃ has a density of 1.4093 g/mL.

[mjc123]

If you had 1000g of water (1 L) and added 360g of solute, the resulting solution would occupy a volume > 1L.
So if 1L of solution weighs 1360g, that means some amount of water < 1000g has been mixed with some amount of solute > 360g. But how much, we don't know without further information.

And by the way, even if it was 1000g water + 360g solute, that wouldn't be 36% solute. It would be 360/1360 = 26.5% solute.

(Borek, OP said "Aq" not "Ag")

[Borek]

(Borek, OP said "Aq" not "Ag")

Oops, q underlined in the subject looks as g, my mistake.

[billnotgatez]

...
And by the way, even if it was 1000g water + 360g solute, that wouldn't be 36% solute. It would be 360/1360 = 26.5% solute.
...

@mjc123
correct me if I am wrong
640 g water + 360 g solute = 1000 g solution
would be 360/1000 = 36%

if the solute was NaCl  sugar then we would have something more than 640 ml of solution.
If the solute was Ethanol then we would have something less than 640 ml of solution.

[billnotgatez]

I made a mistake
I meant water and sugar

[mjc123]

@mjc123
correct me if I am wrong
640 g water + 360 g solute = 1000 g solution
would be 360/1000 = 36%

That is right, but we're told that 1L is 1360g solution. The answer to the question in the title is no.