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?

1 Answer
anor277
Jun 30, 2017

"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.

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