[DukeBlueDvl]

I'm going crazy over this question that I can't complete
The question reads: You prepare a "stock" solution by dissolving 50.0 grams of potassium chloride (KCl) in enough water to make 250mL of solution. If you place 100mL this stock into a beaker and then add (another) 50mL of water to it, what is the resulting molarity?

I figure that I need to find the molarity of the "stock" solution:
My work: Molar Mass of KCl = 74.55g/mol
50g/74.55 = .67moles
250mLx1l/1000mL = .25L ----> 1000/250 = 4
4x.67/.25L = 2.68mol/L

I can't figure out where to go after that. Do I plug the 2.68mol/L to a different equation?

[Yggdrasil]

For dilutions, you can use the following equation:

M₁V₁ = M₂V₂

Where the Ms refer to the molality of the solutions and V refer to the volumes before and after dillution.

So in this case, your M₁ is 2.68M and your V₁ is 100mL.  When you add 50mL of water, the final volume of your solution is (approximately )150mL so V₂ is 150mL.  Simple algebra gets you to the answer of M₂ = 1.79M.

[Borek]

Alternatively think this way: your stock solution contains 50g KCl in 250 mL, you are using 100mL - so you took 100/250*50 = 20g KCl and dissolved it in 150mL (which is the final volume - 100+50).

Both ways are correct, Yggdrasil's is more formal, my probably faster

[DukeBlueDvl]

I just wanted to say a big ole "Thank You" for your help with this equation. It was much appreciated by not only me but by my classmates that shared the same dilemma.