First, we write the reaction to solve this:
Al(NO₃)₃(aq) => Al³⁺(aq) + 3NO₃⁻(aq)
We need to calculate the concentration of the salt:
Molarity = moles of Al(NO₃)₃ / Volume of solution (L)
Molarity = 2.00 moles / 2.00 L = 1.00 mol/L
The salt is completely disociated, so the concentration of NO3- ion is:
Concentration NO3- = 3 x Concentration of the salt = 3 x 1.00 mol/L = 3.00 mol/L
Answer: 3.00 mol/L
