What is the concentration of chloride ions (in) in a solution prepared by adding 0.075 moles of calcium chloride to 425 mL water,assuming no volume change up on mixing?
Thanks!
(CaCl2) = M = moles/L soln.
0.075 mols/0.425 L = ?
Since there are two Cl^- in a mol of CaCl2, the (Cl^-) = twice that.
To find the concentration of chloride ions in the solution, we need to calculate the number of moles of chloride ions present and then divide it by the volume of the solution.
First, let's find the number of moles of chloride ions in 0.075 moles of calcium chloride. Calcium chloride (CaCl2) dissociates into three ions in water: one calcium ion (Ca2+) and two chloride ions (Cl-). So, for every one mole of calcium chloride, we have two moles of chloride ions.
Since we have 0.075 moles of calcium chloride, we have 0.075 moles * 2 = 0.15 moles of chloride ions.
Next, we need to find the volume of the solution. We are given 425 mL of water. Since no volume change occurs upon mixing, the volume of the solution is still 425 mL.
Lastly, we divide the number of moles of chloride ions by the volume of the solution to get the concentration in moles per liter (M).
Concentration (in moles per liter) = (0.15 moles) / (425 mL)
However, to express the answer in moles per liter, we need to convert the volume from milliliters to liters.
1 L = 1000 mL
Concentration (in moles per liter) = (0.15 moles) / (425 mL * 1L/1000 mL)
Concentration (in moles per liter) = (0.15 moles) / (0.425 L)
Concentration (in moles per liter) ≈ 0.353 mol/L
Therefore, the concentration of chloride ions in the solution is approximately 0.353 mol/L.