If 50.0 mL of a 1.00 M solution of CaCl2 (solution A) is mixed with 85.0 mL of pure water to produce solution B and then 42.5 mL of solution B is mixed with 52.0 mL of solution A to give solution C, what is the concentration of CL^-(aq) in the final solution?

Answer 1

from A to B, it was diluted to 1M*50/(50+85)=.37M, but twice that in Cl conc.
then, mixing it with A, (.425*.37*2+.52*1*2)/(.425+.52)=1.43M in Cl

check my thinking.