Assuming the volumes are additive, what is the [Cl-]solution obtained by mixing 210mL of 0.600M KCl and 630ml of 0.385M MgCl2 ?
Responses
chemistry - bobpursley, Friday, September 19, 2008 at 4:20pm
You have to assume each chloride is soluble. How many Cl in each?
Add the Cl, you have the volume, so molar concentration is moles Cl per liter.
THIS IS THE RESPONSE I GOT AND I HAVE NO IDEA WHAT IT MEANS
Have you tried doing what Bob Pursley said? It's a 1, 2, 3 type problem.
By the way, it helps us if you post under the same screen name; otherwise, we get the problems confused as we go from one to the other.
i would but i don't understand
Figure how many moles of Cl are in each volume. Add those moles.
Divide by the total volume.
how would i find the mole of Cl
What's the definition of molarity?
M = #mols/L. So rearrange that, as a formula, to #mols = M x L.
#mols Cl in KCl = M x L.
#mols Cl in MgCl2 = M x L x 2 (because there are 2 mols Cl per mol MgCl2.
Add the mols Cl together.
Then M = total #mols Cl/total volume in liters. Post your work if you get stuck and I'll find the error but I expect you can work through this. Good luck.
To calculate the concentration of chloride ions ([Cl-]) in the solution obtained by mixing KCl and MgCl2, we can use the concept of molarity (M) and the principle of volume additivity.
First, we need to determine the moles of chloride ions in each individual solution. This can be done using the formula:
moles = volume (in liters) x molarity
For the KCl solution:
moles of Cl- in KCl = 0.600 M x 0.210 L = 0.126 moles
For the MgCl2 solution:
moles of Cl- in MgCl2 = 0.385 M x 0.630 L = 0.24255 moles
Next, we can add the moles of chloride ions from both solutions to find the total moles of chloride ions:
total moles of Cl- = moles of Cl- in KCl + moles of Cl- in MgCl2
total moles of Cl- = 0.126 moles + 0.24255 moles
total moles of Cl- = 0.36855 moles
Finally, we need to convert the total moles of chloride ions to the concentration in the final solution. Since the total volume is the sum of the volumes of KCl and MgCl2 solutions, we can use the equation:
concentration = moles of Cl- / total volume (in liters)
total volume = 0.210 L + 0.630 L = 0.840 L
Consequently:
[Cl-]solution = total moles of Cl- / total volume
[Cl-]solution = 0.36855 moles / 0.840 L
[Cl-]solution = 0.4387 M
Therefore, the concentration of chloride ions in the final solution after mixing 210mL of 0.600M KCl and 630mL of 0.385M MgCl2 is approximately 0.4387 M.