If you are making 230.0 mL of 2.70% (m/m) CuSO45 H2O, how many grams of the solute should you use?
Who knows?
Let x = mass solute, then
[x/(230+x)]*100 = 2.70
Solve for x BUT note that the volume certainly will NOT be 230 mL, probably more than 230 mL.
If you want 230 mL of 2.70% (w/v), that is 2.70 x (230/100) = 6.21g CuSO4.5H2O and that WILL be 230 mL but that is w/v% and not w/w%(m/m%).