what volumes of 2M and 4M CaCl2have to be mixed to give 6Liters of 2.5M CaCl2?

The long way of solving this problem is the easier to understand.

moles = M x L
moles of 4M + moles 2M = moles 6L
Let x = volume used from each soln.
(4M*xL) + (2M*xL) = 2.5M x 6L = 15.0 moles.
4x + 2x = 15
Solve for x and I get 2.5L. Check this out this way.
Since moles = M x L and mols of the mixture must = moles of the final solution, then
(2M*2.5L) + (4M*2.5L) = 5moles + 10 moles = 15 moles = what we wanted (6L x 2.5M = 15 moles).

where did you get that 2.5?

Which 2.5. There are two 2.5. One is 2.5M and one is 2.5L.

The 2.5M comes from the 6L of 2.5M solution that you are preparing. The 2.5L is the volume of the 2M solution and the 4M solution that you are to mix together to obtain the 6L of 2.5M solution. HOWEVER, this is not the right solution because it only adds up to 5.0 L and you wanted 6L. I'll leave this up to give you time to read it then I'll erase it and get back to you with the right solution.

Okay. im reading it and trying to get it. Ill wait for your answer! Thanks, tomorrow im having a test on this and im doing some exercises.

I'll wait for the right answer

Second try. How's this?

Let X = volume in liters of the 2M soln.
Then 6-X = volume in liters of the 4M soln.
moles 2M + moles 4M = 15 = moles of the 6L soln (6L x 2.5M = 15.0moles)

2X + 4(6-X) = 15
Solve for X
I get 4.5L of the 2M and 1.5L of the 4M to give 6L of 2.5M

Let me read it.

Is that the right answer now?

I checked and its okay.. thank you

Can you help me out, with another one?

It looks right to me. The problem gives us 2M and 4M CaCl2 and wants 6L of 2.5M CaCl2. If you mix 1.5L of the 4M and 4.5L of the 2M that will give you 6L of 2.5M.

4.5L + 1.5L = 6L so that's ok.
4.5L x 2M = 9 moles + 1.5L x 4M = 6 moles; the total is 15 moles CaCl2 and since M = moles/L then 15 moles/6L = 2.5M, That's what the problem wanted.

Yes but post it as a new question. It may be a little before I can look at it. I have an errand to do but I'll be back in about an hour.