How many milliliters of a 110ml solution of 1.5 m alcl3 must be used to make 70ml of a solution that has a concentration of 0.2m cl–?

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.

The answer is 323 g.