Do you go to UCI by any chance? Anyway, this is how the TA helped me:
6.25% w/vol
1000 g/L
62.5 g/L
150 mg/ 182.2 = 0.82 mmol x 74 (the NaOCl MW) = 60.92 mg
62.5 mg/mL x (x)mL = 60.92 mg
60.92 mg/ 62.5 = 0.97 mL which is roughly 1
and we know that our calculation is correct because the procedure asks for 1.2 mL, so it's fairly close!
Hope that helps!!
Calculate the amount (volume) of 6.25% (wt./vol.) NaOCl solution (commercial bleach)
required to oxidize 150 mg of 9-fluorenol to 9-fluorenone. Whenever appropriate, use
balanced chemical equations as a part of your calculation.
Any help is appreciatede.
2 answers
I go to UCI, and this helped a lot. Thank you!