Question

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

Answer 1

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`