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

Answer 1

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

#"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#

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Answer 2

Jeremiah Adams

The final concentration of the KmnO4 solution after dilution is 0.5 times its initial concentration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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