Assuming the volumes are additive, what is the [Cl-] in a solution obtained by mixing 240ml of 0.635M KCl and 640ml of 0.380M MgCl2 ?
molarity number of moles of substance / volume of soln in litres so number of moles molarity X volume of soln in litres so 240 ml 0.24 L and 640 ml 0.64 L so number of moles of KCl 0.635 X 0.24 0.1524 moles of KCl similarly number of moles of MgCl2 0.38 X 0.64 0.2432 moles of MgCl2 now when 1 molecule of KCl dissociates 1 K+ and 1 Cl- are given out so 1 mole of KCl will give 1 mole of Cl- so 0.1524 moles of KCl will give 0.1524 moles of Cl- but when 1 molecule of MgCl2 dissociates 2 Cl- are given out so 1 mole of MgCl2 will give 2 mole of Cl- so 0.2432 moles of MgCl2 will give 0.2432 X 2 0.4864 moles of Cl- now total number of moles of Cl- 0.1524 + 0.4864 0.6388 moles of Cl- and total volume 0.24 + 0.64 0.88 L (as volumes are additive given in question) so [Cl-] or molarity 0.6388/0.88 0.7259 M please check the maths
Mol KCl in 240mL of 0.635M solution 240/1000*0.635 0.1524 mol KCl KCl dissociates: KCl ? K+ + Cl- Therefore you have 0.1524 mol Cl- ions Mol MgCl2 in 640mL of 0.380M solution 640/1000*0.380 0.2432 mol MgCl2 MgCl2 dissociates : MgCl2 ? Mg 2+ + 2Cl- Therefore you have 0.2432*2 0.4864 mol Cl- Total mol Cl- 0.1524 + 0.4864 0.6388 mol Total volume 240+640 880mL 0.880L Molarity 0.6388/0.880 0.726 M solution of Cl- ions.
