Question

If there’s 340mL of a 0.5 M NaBr solution, what will the concentration be if 560 mL of water are added?

Answer 1

Around `0.19 \ "M"`.

Explanation:

In a `340 \ "mL"=0.34 \ "L"` of `0.5 \ "M"` of sodium bromide solution, there exist:

`0.34color(red)cancelcolor(black)"L"*(0.5 \ "mol")/(color(red)cancelcolor(black)"L")=0.17 \ "mol"` of sodium bromide

When we add `560 \ "mL"=0.56 \ "L"` of water, then the resulting volume is:

`0.34 \ "L"+0.56 \ "L"=0.9 \ "L"`

And so, the new concentration is:

`c=n/v=(0.17 \ "mol")/(0.9 \ "L")~~0.19 \ "mol/L"`

`=0.19 \ "M"`