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?
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. To find the molarity of the resulting solution, we can use the dilution formula:
M1V1 = M2V2
where:
M1 = initial molarity of the solution (0.50 M)
V1 = initial volume of the solution (75 mL)
M2 = final molarity of the solution (unknown)
V2 = final volume of the solution (375 mL)
Now let's solve for M2:
M2 = (M1V1) / V2
M2 = (0.50 M * 75 mL) / 375 mL
M2 = 0.10 M
Therefore, the molarity of the resulting solution will be 0.10 M.
2. To find the final volume of the solution, we can rearrange the dilution formula:
M1V1 = M2V2
where:
M1 = initial molarity of the solution (1.38 M)
V1 = initial volume of the solution (250 mL)
M2 = final molarity of the solution (0.30 M)
V2 = final volume of the solution (unknown)
Now let's solve for V2:
V2 = (M1V1) / M2
V2 = (1.38 M * 250 mL) / 0.30 M
V2 = 1150 mL
Therefore, the final volume of the solution will be 1150 mL.
3. To find the initial mL volume of HCl needed, we can use the dilution formula:
M1V1 = M2V2
where:
M1 = initial molarity of the solution (1.00 M)
V1 = initial volume of the solution (unknown)
M2 = final molarity of the solution (0.15 M)
V2 = final volume of the solution (500 mL)
Now let's solve for V1:
V1 = (M2V2) / M1
V1 = (0.15 M * 500 mL) / 1.00 M
V1 = 75 mL
Therefore, the initial mL volume of HCl needed will be 75 mL.
4. To find the final molarity of the solution, we can use the dilution formula:
M1V1 = M2V2
where:
M1 = initial molarity of the solution (0.93 M)
V1 = initial volume of the solution (183 mL)
M2 = final molarity of the solution (unknown)
V2 = final volume of the solution (482 mL)
Now let's solve for M2:
M2 = (M1V1) / V2
M2 = (0.93 M * 183 mL) / 482 mL
M2 = 0.35 M
Therefore, the final molarity of the solution will be 0.35 M.
5. To find the final volume of the 0.84 M solution, we can rearrange the dilution formula:
M1V1 = M2V2
where:
M1 = initial molarity of the solution (unknown)
V1 = initial volume of the solution (3.29 L)
M2 = final molarity of the solution (0.84 M)
V2 = final volume of the solution (unknown)
Now let's solve for V2:
V2 = (M1V1) / M2
V2 = (15.3 moles / 9.21 L) * (0.84 M / 3.29 L)
V2 = 3.14 L
Therefore, the final volume of the 0.84 M solution will be 3.14 L.
1. To find the molarity of the resulting solution, we can use the formula: M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
In this case, M1 = 0.50 M, V1 = 75 mL, V2 = 375 mL, and M2 is what we need to find.
Rearranging the formula gives us M2 = (M1 * V1) / V2.
Substituting the values gives M2 = (0.50 M * 75 mL) / 375 mL = 0.10 M.
Therefore, the molarity of the resulting solution will be 0.10 M.
2. To find the final volume of the solution, we can again use the formula: M1V1 = M2V2.
In this case, M1 = 1.38 M, V1 = 250 mL, M2 = 0.30 M, and we need to find V2.
Rearranging the formula gives V2 = (M1 * V1) / M2.
Substituting the values gives V2 = (1.38 M * 250 mL) / 0.30 M = 1,150 mL.
Therefore, the final volume of the solution will be 1,150 mL.
3. In this question, we know the final volume (V2) of the solution (500 mL), the final molarity (M2) of the solution (0.15 M), and the stock concentration (M1 = 1.00 M). We need to find the initial volume (V1) of the stock solution.
Using the formula M1V1 = M2V2, we can rearrange it to V1 = (M2 * V2) / M1.
Substituting the values gives V1 = (0.15 M * 500 mL) / 1.00 M = 75 mL.
Therefore, you would need 75 mL of the stock solution to make a 500 mL solution with a concentration of 0.15 M.
4. To find the final molarity of the LiBr solution, we can use the formula: M1V1 = M2V2.
In this case, V1 = 183 mL, M1 = 0.93 M, V2 = 482 mL, and we need to find M2.
Rearranging the formula gives M2 = (M1 * V1) / V2.
Substituting the values gives M2 = (0.93 M * 183 mL) / 482 mL = 0.35 M.
Therefore, the final molarity of the solution will be 0.35 M.
5. In this question, we want to find the final volume of the 0.84 M solution, given that we started with a stock solution of 15.3 moles of BeI2 and 9.21 L of water, and we took 3.29 L of that solution.
First, we need to calculate the moles of BeI2 in the 3.29 L solution. Using the formula n = c * V, where n is the moles, c is the molarity, and V is the volume in Liters, we can calculate n = 0.84 M * 3.29 L = 2.76 moles of BeI2.
To determine the final volume of the 0.84 M solution, we can rearrange the formula M1V1 = M2V2, where M1 = initial molarity, V1 = initial volume, M2 = final molarity, and V2 = final volume.
In this case, M1 = 15.3 moles / 9.21 L = 1.66 M (concentration of the stock solution), V1 = 3.29 L (volume taken), M2 = 0.84 M (desired final molarity), and we need to find V2.
Rearranging the formula gives V2 = (M1 * V1) / M2.
Substituting the values gives V2 = (1.66 M * 3.29 L) / 0.84 M = 6.46 L.
Therefore, the final volume of the 0.84 M solution will be 6.46 Liters.
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