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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