Question

A `75*mL` volume of `0.211*mol*L^-1` `NaOH` was diluted to `125*mL`. What is the new concentration?

Answer 1

Approx. `0.127*mol*L^-1`..............with respect to `NaOH`

Explanation:

`"Concentration"="Moles of solute"/"Volume of solution"`

And this is all the equation we need for these sort of problems:

`i.e. "moles of solute"="concentration"xx"volume"`

And so.........

`"Concentration of NaOH"=(75xx10^-3*cancelLxx0.211*mol*cancel(L^-1))/(125xx10^-3*L)=??*mol*L^-1.`