Find the amount concentration of the anion in each of the following solutions:

A) 0.50 mol/L barium chloride, BaCl2
B) 6.0 mol/L potassium hydroxide, KOH
C) 0.10 mol/L aluminium chlorate, Al(ClO3)3

Find the concentration of and aluminum and chlorate ions in 0.10 mol/L aluminum chlorate solution, Al(ClO3)3

To find the amount concentration of an anion in a solution, we need to determine the number of moles of the anion present in one liter of the solution.

Let's take a look at each case:

A) 0.50 mol/L barium chloride, BaCl2:
Barium chloride, BaCl2, dissociates into two chloride ions, Cl-. Therefore, each mole of barium chloride will give us two moles of chloride ions.
Since the concentration is given as 0.50 mol/L, this means there are 0.50 moles of barium chloride per liter of solution. As a result, we will have 2 * 0.50 = 1.00 mol/L of chloride ions in the solution.

B) 6.0 mol/L potassium hydroxide, KOH:
Potassium hydroxide, KOH, dissociates into one hydroxide ion, OH-. Thus, each mole of potassium hydroxide will give us one mole of hydroxide ions.
Given that the concentration is 6.0 mol/L, this means there are 6.0 moles of potassium hydroxide per liter of solution. Consequently, we will have 6.0 mol/L of hydroxide ions in the solution.

C) 0.10 mol/L aluminum chlorate, Al(ClO3)3:
Aluminum chlorate, Al(ClO3)3, dissociates into three chloride ions, Cl-. Therefore, each mole of aluminum chlorate will yield three moles of chloride ions.
The concentration provided is 0.10 mol/L, indicating that there are 0.10 moles of aluminum chlorate per liter of solution. This implies that we will have 3 * 0.10 = 0.30 mol/L of chloride ions in the solution.

In summary, the amount concentration of the anion in each solution is:
A) 1.00 mol/L chloride ions
B) 6.0 mol/L hydroxide ions
C) 0.30 mol/L chloride ions