Question

If I have 340 mL of a 0.5 M `NaBr` solution, what will the concentration be if l add 560 mL more water to it?

Answer 1

Approx. `0.20` `mol*L^-1` with respect to `NaBr`

Explanation:

`"Concentration " = " Moles of Solute"/"Volume of Solution"`

`=` `(340xx10^-3*Lxx0.5*mol*L^-1)/((340+560)xx10^-3L)`

The numerator gives the amount of `NaBr` in moles in the initial volume. This is divided again by the new volume of the solution `900xx10^-3*L`.

`=` `??*mol*L^-1?`