Chemistry Forum

How can chloroauric acid be prepared from a piece of gold weighing 0.7 grams? Then I want to prepare a 0.01 molar solution of chloroauric acid pls help me

Dissolve in aqua Regina.

ChenBeier wrote: ↑Sun Feb 11, 2024 2:52 am Dissolve in aqua Regina.

"Do you have specific ingredients carry out this?"
If you don't mind pls
by details

I thing to figure out what is aqua Regina and the calculati8n you have to do by yourself.

Calculate the moles of Gold you have then covert this to the volume of the acid to get 0,01 M

ChenBeier wrote: ↑Sun Feb 11, 2024 3:40 am I thing to figure out what is aqua Regina and the calculati8n you have to do by yourself.

Calculate the moles of Gold you have then covert this to the volume of the acid to get 0,01 M

Great, with the given molar mass of chloroauric acid (\(HAuCl_4\)) being 339.785 g/mol, let's recalculate the moles of gold and then determine the volume of the acid needed for a 0.01 M solution.

1. **Calculate moles of gold:**
\[ \text{Moles of Gold} = \frac{\text{Mass of Gold}}{\text{Molar Mass of Gold}} \]

Given:
\[ \text{Mass of Gold} = 0.7 \, \text{g} \]
\[ \text{Molar Mass of Gold} = 197.0 \, \text{g/mol} \]

\[ \text{Moles of Gold} = \frac{0.7 \, \text{g}}{197.0 \, \text{g/mol}} \]

\[ \text{Moles of Gold} \approx 0.00355 \, \text{mol} \]

2. **Convert moles to volume of 0.01 M solution:**
Using the molar mass of chloroauric acid (\(HAuCl_4\)), the moles of gold can be equated to the moles of chloroauric acid.

\[ \text{Moles of Chloroauric Acid} = 0.00355 \, \text{mol} \]

Now, to find the volume of a 0.01 M solution, you can use the formula:

\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]

Given:
\[ \text{Molarity} = 0.01 \, \text{M} \]

\[ \text{Volume (L)} = \frac{0.00355 \, \text{mol}}{0.01 \, \text{M}} \]

\[ \text{Volume (L)} = 0.355 \, \text{L} \]

So, you would need approximately 0.355 liters (355 milliliters) of the chloroauric acid solution (0.01 M) to prepare it using 0.7 grams of gold.

That's right?

You need about 355 ml solvent for the 0.7 g.

But read how to make it

What is aqua Regina it is 3 part HCl and one part HNO₃ .

https://en.m.wikipedia.org/wiki/Chloroauric_acid
You have to do boiling until all HNO₃ is evaporated as NO₂, later dissolve the product again in conc. HCl to get 355 ml

ChenBeier wrote: ↑Sun Feb 11, 2024 3:56 am You need about 355 ml solvent for the 0.7 g.

But read how to make it

What is aqua Regina it is 3 part HCl and one part HNO₃ .

https://en.m.wikipedia.org/wiki/Chloroauric_acid
You have to do boiling until all HNO₃ is evaporated as NO₂, later dissolve the product again in conc. HCl to get 355 ml

I can't find enough words to thank you
thank you so much