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
6 answers
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.
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.
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.
3 answers