I don't know to agree or disagree about the volume since you don't say what you are assuming about the volume.
0.0510 M means 0.0510 moles/L soln.
0.0510 mol x 19 g F^-/mol = about 0.969 g/L soln.
Since the density of the soln is 1.00 g/mL, then this is 0.969 g F^-/1000 g soln
% w/w = (0.969g/1000g)*100 = 0.0969%.
For ppm, I find the factor 1 ppm = 1 mg/L to be useful.
0.000969 g/mL x (1000mg/g) = 0.969 mg/mL x (1000 mL/L) = 969 g/L = 969 ppm.
Check all of this carefully.
Express the concentration of a 0.0510 M aqueous solution of fluoride, F-, in mass percentage and in parts per million. Assume the desnity of the solution is 1.00 g/mL
Since, we're assuming the density I think that also means were assuming the volume.. but where would you go from there? please help!
2 answers
yes that was correct! thank you
1 answer