Too many questions on one post.
1.
0.50M x (75 mL/375 mL) = ? if you can make a 0.50M CaF2 solution. I didn't think it was that soluble.
2.
Use the dilution formula.
mL1 x M1 = mL2 x M2
1. If I dilute 75 mL of 0.50 M CaF2 solution to a final volume of 375 mL, what will be the molarity of the resulting solution?
2. What will be the final volume of the solution be if I start with a 250 mL of 1.38 M MgCl2 solution to make a solution with a concentration of 0.30 M?
3. Find the initial mL volume of HCl I would need to make a 500 mL of a 0.15 M HCl solution from a stock bottle of 1.00 M HCl.
4. The final volume of my solution of LiBr must be 482 mL. If I started with 183 mL of a 0.93 M LiBr solution, what will the final molarity of the solution be?
5. I create a stock solution using 15.3 moles of BeI2 and 9.21 Liters of water. I then take 3.29 L of that solution and need to dilute it to a molarity of 0.84 M. What will be the final volume, in Liters, of the 0.84 M solution?
2 answers
You are right DrBob, the maximum solubility of CaF2 at 25-deg C = 2.04 x 10^-3M. The problem data is ~2,500X more soluble than what can be achieved.
1 answer