Answer:
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
Step-by-step explanation:
Step 1: Data given
2.0 M Al(NO3)3
Step 2:
Al(NO3)3 in water will dissociate as following:
Al(NO3)3 → Al^3+ + 3NO3^-
For 1 mol of Al(NO3)3 we will have 1 mol of Al^3+ and 3 moles of NO3^-
We know that the molarity of Al(NO3)3 = 2.0 M, this means 2.0 mol/ L
The mol ratio Al(NO3)3 and Al^3+ is 1:1 so the molarity of Al^3+ is 2.0 M
The mol ratio Al(NO3)3 and NO3^- is 1:3 so the molarity of NO3^- is 6.0M
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)