To what volume should you dilute 131 mL of an 8.20 M CuCl2 solution so that 49.0 mL of the diluted solution contains 4.53g CuCl2 ?

Chemistry...Help!! - Dr Russ, Wednesday, December 8, 2010 at 4:43am
Molar mass of CuCl2 is 134.5 g mol^-1

so solution needs to contain

4.53 g / 134.5 g mol^-1
=0.03368 mole

concentration is

0.049 L/0.03368 mole
=1.455 molar

thus we need to dilute the 8.20 M solution by a factor
8.20 M/1.455 M= 5.636

so the new total volume is

131 ml x 5.636
=738.4 ml

check on the concentration

old solution contains
131 x 8.20 mmoles = 1074 mmoles

new solution contains
1.455 x 738.4 mmole = 1074 mmole


Chemistry...Help!! - Lisa, Help needed!, Wednesday, December 8, 2010 at 1:59pm

Thank you for all the detail, but i'm still confused......so 738.4mL is the answer?

It doesn't help us help you if you simply say you are confused without telling us what you understand, what you don't understand, and if possible, why you don't understand.

the second part is moles/volume not volume/moles

Yes, 738.4 mL is the answer to the question. Here is the explanation on how to arrive at that answer:

1. Calculate the number of moles of CuCl2 in the 4.53g sample using the molar mass of CuCl2. The molar mass is 134.5 g/mol, so 4.53 g / 134.5 g/mol = 0.03368 moles.

2. Determine the concentration of CuCl2 in the diluted solution. You have 49.0 mL of the diluted solution, which is equal to 0.049 L. Divide the volume by the number of moles: 0.049 L / 0.03368 moles = 1.455 M.

3. Find the dilution factor by dividing the initial concentration (8.20 M) by the desired concentration (1.455 M): 8.20 M / 1.455 M = 5.636.

4. Multiply the initial volume (131 mL) by the dilution factor to get the new total volume: 131 mL x 5.636 = 738.4 mL.

So, to dilute the 131 mL of the 8.20 M CuCl2 solution, you need to add enough solvent to make a final volume of 738.4 mL in order to achieve a concentration of 1.455 M in 49.0 mL.