To what volume should you dilute 124mL of an 8.10M CuCl2 solution so that 49.0mL of the diluted solution contains 4.81g CuCl2?
Molar Mass of CuCl2= 134.45 g/mol
Convert: 4.81g * (1mol/ 13.45g CuCl2)= 0.0358mol CuCl2
[CuCl2]= M= mol/L= (0.0358mol/ 49ml) * (1000mL/1L)= 0.7306M
Dilute the 8.20M of solution by a factor of 0.7306M:
8.20M/ 0.7306M= 11.22
Total Volume= 124mL * 11.22= 1391.73mL
----Where do I go from here to get the final answer?---
I think you're there if you correct the typo in your next to last step. That 8.20M should be 8.10 M so you get a slightly different factor and that x 124 mL will give you the right answer.
To get the final answer, you need to convert the total volume from milliliters to liters.
1391.73 mL * (1 L/ 1000 mL) = 1.39173 L
Therefore, you need to dilute 124 mL of an 8.10 M CuCl2 solution to a total volume of 1.39173 L.
To find the volume to which you should dilute the 124mL of the 8.10M CuCl2 solution, given that 49.0mL of the diluted solution contains 4.81g CuCl2, you can use the formula for dilution:
C1V1 = C2V2
where C1 is the initial concentration (8.10M), V1 is the initial volume (124mL), C2 is the final concentration (0.7306M), and V2 is the final volume (which we are trying to find).
Rearranging the equation, we have:
V2 = (C1 * V1) / C2
Plugging in the values, we have:
V2 = (8.10M * 124mL) / 0.7306M
= (8.10 * 0.124) / 0.7306
= 1.38 L
So, you should dilute the 124mL of the 8.10M CuCl2 solution to a final volume of 1.38 L in order for 49.0mL of the diluted solution to contain 4.81g CuCl2.