Question

How do I calculate the following quantity? Volume of 2.106 M copper (II) nitrate that must be diluted with water to prepare 670.2 mL of a 0.8351 M solution?

Answer 1

`"Volume"="Moles"/"Concentration"=266*mL`

Explanation:

We need `670.2*mLxx10^-3*mL*L^-1xx0.8351*mol*L^-1`

`=0.660*mol` `"copper nitrate"`.

We have a `2.106*mol*L^-1` solution of `Cu(NO_3)_2(aq)` available....

And thus we take the quotient, `"volume"=(0.660*mol)/(2.106*mol*L^-1)` `=` `0.266*L=265.7*mL`, and then dilute this volume up to `670.2*mL`.

You would usually never dilute solutions like this, so this question is not very practical and does not reflect standard lab practice. Ordinarily we would take `100*mL` volumes or so of a mother solution, and dilute appropriately by a factor of 2 or 10.....

The important relationship was.....

`"Concentration"="Moles of solute"/"Volume of solution"`,

i.e. `C("concentration")=(n("moles"))/(V("Litres"))`, and thus `n=CxxV`, `V=n/C` etc.