how can I make gold chloride from colloidal gold

[13ishmael]

:lol:
Guys I do not wish to delve into alchemy. I found the answer I was looking for.
You guys are the best. Thanks for all of your help. I really do appreciate it.

 

[Marcel]

If you reduce the amount of water, colloidal gold should form bigger molecules. So either boil it down or better destill it, so you wont loose so much in the process. The suggested methods (adding sulphuric acid and boil) may result in the same effect but I dont think you need the sulphuric acid if the collidale solution is free of base metals and/or their salts.

Btw: Adding gold chloride to the collidale solution and stirring it should also work. The mechanism is to increase the concentration of the particles so they adhere to each other.

 

[goldsilverpro]

So how big will the gold metal big once it is precipitated

50ppm = .05g/liter pure Au. You have .473 liters. Therefore, you have .05 X .473 = 0.0237g Au

For that quantity, I figured that a gold BB of that weight would have a diameter of about 1/20". or .052" or .133cm. If it was much smaller, it would be hard to pick it up between thumb and forefinger without losing it.

V(sphere) = (4/3) X pi X r³
Eq#1 = r³ = (3 X V)/(4 X pi)

V of .0237g Au = .0237/19.32 = .00123cm³

Therefore, from Eq#1 above: r³ = (3 X V)/(4 X pi) = (3 X .00123)/(4 X pi) = .000294cm³
Therefore, from my calculator: the cube root of .000294 = r = .0665cm
Therefore, the diameter = 2r = .0665 X 2 = .133cm = .133/2.54 = .052" = about 1/20"

All the math assumes that the solution actually does contain 50ppm of gold, which, I feel, is somewhat doubtful.