Answer to Question #56284 in General Chemistry for Emily
How many equivalents of chloride ion are in 20.0 mL of a 2.0 M solution of CaCl2 and how do you calculate this?
СаCl2 in aqueous solution dissociates according to the equation: СаCl2 = Ca2+ + 2Cl– Thus, 1 molecule СаCl2 forms a 2 ion Cl– 1 liter (1000 ml) of 2.0 M solution СаCl2 contains 2 mol СаCl2 20.0 ml of 2.0 M solution СаCl2 contain 2*20/1000 mol = 0.04 mol СаCl2. From 0,04 mol СаCl2 formed 2*0,04 = 0,08 equivalent of the chloride ions
Answer: Formed 0.08 per equivalent of chloride ions