Asked by Joy
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!
Since, we're assuming the density I think that also means were assuming the volume.. but where would you go from there? please help!
Answers
Answered by
DrBob222
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.
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.
Answered by
Joy
yes that was correct! thank you
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.