A solution contains an unknown mass of dissolved silver ions. When potassium chloride is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 245 mg. What mass of silver was in the original solution? (Assume that all of the silver was precipitated out of solution by the reaction.)

To determine the mass of silver in the original solution, we can use the information provided about the white precipitate formed when potassium chloride is added.

1. Determine the molar mass of silver chloride (AgCl):
- The molar mass of silver (Ag) is 107.87 g/mol.
- The molar mass of chlorine (Cl) is 35.45 g/mol.
- Therefore, the molar mass of silver chloride (AgCl) is 107.87 g/mol + 35.45 g/mol = 143.32 g/mol.

2. Calculate the number of moles of silver chloride precipitated:
- Given that the mass of the precipitate is 245 mg (or 0.245 g).
- Use the equation: Moles = Mass / Molar mass.
- Moles of silver chloride = 0.245 g / 143.32 g/mol.
- Moles of silver chloride = 0.0017082 mol.

3. Determine the molar ratio between silver ions (Ag+) and silver chloride (AgCl):
- According to the balanced chemical equation, 1 molecule of silver chloride (AgCl) forms for every 1 silver ion (Ag+).
- Therefore, the molar ratio between silver ions and silver chloride is 1:1.

4. Determine the number of moles of silver ions in the original solution:
- Since the molar ratio between silver ions and silver chloride is 1:1, the number of moles of silver ions is also 0.0017082 mol.

5. Calculate the mass of silver in the original solution:
- Given that the molar mass of silver (Ag) is 107.87 g/mol.
- Use the equation: Mass = Moles × Molar mass.
- Mass of silver in the original solution = 0.0017082 mol × 107.87 g/mol.
- Mass of silver in the original solution = 0.1847 g.

Therefore, the mass of silver in the original solution was 0.1847 grams (or 184.7 mg).

To find the mass of silver in the original solution, we need to consider the reaction between silver ions and potassium chloride that formed the white precipitate. The chemical equation for this reaction is:

Ag+(aq) + Cl-(aq) → AgCl(s)

From the equation, we can see that 1 mole of silver ions (Ag+) reacts with 1 mole of chloride ions (Cl-) to form 1 mole of silver chloride (AgCl).

First, let's determine the molar mass of silver chloride (AgCl). The molar mass of silver (Ag) is approximately 107.87 g/mol, and the molar mass of chlorine (Cl) is approximately 35.45 g/mol.

So, the molar mass of AgCl = molar mass of Ag + molar mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol

Next, we'll calculate the number of moles of silver chloride using the given mass of the precipitate.

Number of moles of AgCl = mass of AgCl / molar mass of AgCl

The mass of the AgCl precipitate is given as 245 mg. To convert this to grams, divide by 1000.

Mass of AgCl = 245 mg / 1000
= 0.245 g

Number of moles of AgCl = 0.245 g / 143.32 g/mol

Now, since the reaction is 1:1 between silver ions and silver chloride, the number of moles of silver ions (Ag+) is equal to the number of moles of silver chloride (AgCl).

Number of moles of Ag+ = Number of moles of AgCl

Finally, we can calculate the mass of silver (Ag) in the original solution. Since the molar mass of silver (Ag) is approximately 107.87 g/mol, we can use the number of moles of silver ions (Ag+) to find the mass.

Mass of Ag = Number of moles of Ag+ × molar mass of Ag

So, the mass of silver in the original solution is:

Mass of Ag = Number of moles of Ag+ × molar mass of Ag
= Number of moles of AgCl × molar mass of Ag
= 0.245 g / 143.32 g/mol × 107.87 g/mol

Calculating this will give the mass of silver in the original solution.

245 mg = 0.245 grams.

moles AgCl = grams/molar mass AgCl.

Mols Ag must be the same as moles AgCl. Convert moles Ag to grams Ag. g = moles x molar mass.

0.245