What is the final concentration of a #25.7*mL# volume of #KmnO_4(aq)# that is diluted to a #50.00*mL# volume?

1 Answer
Sep 21, 2016

Answer:

Approx. #1.56*mol*L^-1#, as you have diluted the starting solution by half.

Explanation:

#"Concentration"# #=# #"Moles of solute"/"Volume of solution (L)"#, and thus #"Concentration"# has the units #mol*L^-1#. With this relationship we can solve most problems of dilution.

We started with #25.7*mL# of a #3.12*mol*L^-1# #KMnO_4# solution.

#"Moles of "KMnO_4# #=# #25.7xx10^-3*Lxx3.12*mol*L^-1# #=# #??*mol#.

This molar quantity was then diluted to a #50.00*mL# volume.

#"Final concentration"# #=# #(25.7xx10^-3*cancelLxx3.12*mol*cancel(L^-1))/(50.00xx10^-3*L)# #=# #??mol*L^-1#