how would you prepare 250 mL of 0.69M Al2(SO4)3 solution from a 2.0M Al2(SO4)3 stock solution?
Ah Ha! So you're posting the problem right. And you see there is a difference in m and M.
How many mols do you want? That's M x L = mols.
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
To prepare the desired 0.69M Al2(SO4)3 solution from a 2.0M Al2(SO4)3 stock solution, you need to calculate the volume of the stock solution and the volume of water required.
Let's break down the steps:
1. Use the formula for dilution: C1V1 = C2V2
- C1 is the concentration of the stock solution (2.0M)
- V1 is the volume of the stock solution we need to find
- C2 is the desired concentration (0.69M)
- V2 is the final volume of the solution (250 mL or 0.250L)
2. Rearrange the formula to solve for V1:
- V1 = (C2 * V2) / C1
3. Substitute the known values:
- V1 = (0.69M * 0.250L) / 2.0M
- V1 = 0.08625L or 86.25 mL (rounded to 2 decimal places)
So, you need to measure 86.25 mL of the 2.0M Al2(SO4)3 stock solution and then add water to make a total volume of 250 mL.