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The solution of VmLAl 2 (SO 4) 3 contains agSO 4 2-, and its mass is ag 96g/mol = a 96 mol.

Then the amount of aluminum ion is a 96 mol × 2 3 = a 48×3 mol,

If half of the solution is diluted to 2ml (2v1000l) by adding water,

Then the concentration of Al 3+ in the diluted solution is a 48× 3mol×1.22v1.000l =1.25a72vmol/L. 。

So the answer is: 125A 72V.