What volume OF SOLVENT must we add to a #500*mL# solution of #1.2*mol*L^-1# concentration to dilute the concentration to #1.0*mol*L^-1#?

1 Answer
Mar 2, 2017

Answer:

Look at this dimensionally...........we must add #100*mL# to the initial volume, to make up to a volume of #600*mL#...........

Explanation:

#C_1V_1=C_2V_2#, if we multiply the DIMENSIONS of concentration and volume, i.e.:

#"Concentration"xx"Volume"-=mol*cancel(L^-1)xxcancelL=mol#, i.e. #CV-="moles, i.e. amount of substance. "#

And so, we solve for #V_2=(C_1V_1)/C_2# (and CLEARLY this has the units of volume, as we require.)

And so..........,

#V_2=(1.2*cancel(mol)*cancel(L^-1)xx500*cancel(mL)xx10^-3*L*cancel(mL^-1))/(1.0*cancel(mol)*cancel(L^-1))#

#=0.60*L#

And so we ADD #100*mL# to the original solution.........

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