text: need help? when 35 ml of a 125 ml stock solution containing 1.1 m alcl3 is diluted to 61 ml, what is the concentration of cl- in the new solution? number of chloride mol = c(v/1000) x 3                                        = 1.1 x 35/1000 x 3                                        = 0.1155 chloride mol. [chloride ion] = 0.1155/ (61/1000)                       = 1.893442623m  hope it's right.