When AlCl3 is in aqueous form, it dissociates into ions of Al3+ and Cl-. It's dissociation equation is as follows:
AlCl3 = Al3+ + 3Cl-
We are given all needed information for the calculation of the volume needed to have a 0.2 M Cl-. We calculate as follows:
0.2 M Cl- ( .070 L ) = 0.014 Cl- = 4.67x10^-3 mol AlCl3
4.67x10^-3 / 1.5 = 3.11x10^-3 L = 3.11 L of the 1.5 M AlCl3 solution is needed
Hope this answers the question.
