Question

if 2.00 moles of aluminum nitrate are dissolved to form 2.00l of solution , the concentration of the NO3 ion will be ?

Answer 1

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