Calculating how much solvent is needed to dilute 1.5M of Cu(NO3)2

[Cee]

Hello,

I just want to make sure that my calculations are correct before doing an experiment. I need to dilute 250ml of 1.5M Cu(NO₃)₂ to 0.5M and distilled water is to be used as the solvent. So I'm not sure how much (volume) of distilled water need to be added.

I used the equation C1xV1=C2xV2
where C1=1.5M
V1=250mL
C2=0.5M
V2=?
so

1.5x250=0.5xV2
V2=750mL

I'm not sure if I'm doing it right because I'm not sure if the 750mL is the volume of the solvent that is to be added to the 250mL of Cu(NO₃)₂ to make the concentration 0.5M or if 750mL is the final volume of the Cu(NO₃)₂...

I really need to know how to change the concentration, any help would be greatly appreciated please?

 

[CompuChip]

The big M stands for mol/L. As you probably know, a mole is a measure of the number of molecules, so the concentration in M basically tells you something about the number of molecules per liter of solution. So initially you have 250 mL of 1.5 mol/L solution, which means that there is 0.250 L x 1.5 mol/L = 0.375 mol of Cu(NO₃)₂ in your bottle (note how I converted the volume from mL to L).

You need to dilute this to 0.5 mol/L. So how much solvent should you have in total? How much do you need to add to the 250 mL you had?

Now if you look back what you've done, you should see the C1 x V1 = C2 x V2, but hopefully you better understand what it means.

 

[Borek]

To add to CompuChip answer:

I used the equation C1xV1=C2xV2

This is just mass balance (or mass conservation) - amount of substance in the solution after dilution is identical to amount of substance before the dilution. That means both V₁ & V₂ refer to the total volumes.