Dear Harsh Kumar , I hope you have already solved the question, but if not may be this could help you:

The methodology described by the authors means that they used 50 ml of cell supernatant as the solvent for the hydrogen tetrachloroaurate, so this is the volume you have to take into account together with the final concentration (3 mM) to obtain the number of moles of hydrogen tetrachloroaurate you need and from this value you can calculate the amount of hydrogen tetrachloroaurate you need to weight to add to your cell supernatant.

Molarity=moles of substance/liters of final solution

mM=moles of substance /mL of final solution

1mM=0.001M= moles hydrogen tetrachloroaurate/50mL=0.05L

moles of hydrogen tetrachloroaurate=0.001x0.05=0.00005moles of hydrogen tetrachloroaurate

Each mol of hydrogen tetrachloroaurate has 1 mol of Au, so you need as much moles of Au as tetrachloroaurate.

Now you need the molecular weight of your hydrogen tetrachloroaurate. Remember, it be hydrated with 3 or 4 water molecules, so you need to now this to get the molecular weight.

393.833 g/mol (trihydrate) 411.85 g/mol (tetrahydrate)

Supposse you have the trihydrate, and that it is 99.99% pure:

0.00005moles x393.833g/mol=0.0197g.

So you would need 20mg to mix with your 50mL of cell supernatant to have a 1mM Au solution.

Hope it helps

Harsh Kumar

How much gold solution depends on how much gold particles you wish to produce. Manuel Gómez has explained it well above. We usually formulate chloroauric acid as HAuCl4.xH2O and even though there's a constant ratio (or stoichiometry) between the acid precursor and gold, the other complication is that chloroauric acid is extremely hygroscopic and x changes in a normal atmosphere where humidity/water is present. Working in a dry glove box can eliminate this problem.

interested Q really