Which of the following best describe the particles present in a 2.0 m aqueous solution of al(no3)3? 2 m al(no3)3(aq)2 m al3+(aq) and 6 m no3-(aq)2 m al3+(aq) and 3 m no3-(aq)2 m al3+(aq) and 18 m no3-(aq)

2,0 M of Al³⁺ and 6,0M of NO₃⁻.

Explanation:

In water, Al(NO₃)₃ dissociates as follows:

Al(NO₃)₃ → Al³⁺ + 3 NO₃⁻

That means in water, you will have 1 mole of Al³⁺ and 3 moles of NO₃⁻ per mole of Al(NO₃)₃.

Now, as the solution of Al(NO₃)₃ is 2,0M, you will have

2,0 M of Al³⁺ and 6,0M of NO₃⁻.

I hope it helps!

b. is the answer hoped i        

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole