Answer #7 and show work please. Also I posted 6 and the answer I got was complet
ID: 695562 • Letter: A
Question
Answer #7 and show work please. Also I posted 6 and the answer I got was completely wrong with the poster saying more polar solvents travel further on tlc... for 6C would fluorenone not be in the initial fractions because it is a polar molecule being eluted with a nonpolar solvent or am I wrong? Thank you. 6. The synthesis of fluorenone dichromate in acetic acid. mp 83'C) was attempted by oxidizing fluorene (mp 114'C) with sodium fluorenone leorene NaCr 0 acetic acid Upon heating, the mixture was diluted with water and a crude product was isolated by filtration and allowed to dry. This product was then analyzed by melting point determination (mp 72-78'C) and TLC using silica gel plates developed with 20% dichloromethane in hexane (shown below). a) (4 pts) Calculate the Rr values for spots A and B. solvent front - B. origin Rf (B) = (4 pts) Briefly describe the composition of the product. Rationalize both the mp and TLC data in your response. b) (4 pts) Column chromatography using alumina was carried out on the product by eluting with ligroin (a mixture of pentane isomers). Would you expect to collect fluorenone in one of the initial fractions? Briefly explain your answer c) 7. (4 pts) A student has 28.8 grams of sodium benzoate (FW: 144.10 g/mol) and wishes to perform an extraction with ether in order to isolate benzoic acid. To the nearest milliliter (ensuring that all of the sodium benzoate is converted to benzoic acid) how much 6.0 M HCI does the student have to add? a) 2 mL b) 12 mL c) 30 mL d) 34 mL e) 60 mL f) None of these
Explanation / Answer
7.
Moles of sodium benzoate (C6H5COONa) = mass of sodium benzoate / molar mass of sodium benzoate
= 28.8 g/ (144.10 g/mol)
= 0.200 mol
Reaction: C6H5COONa + HCl --------> C6H5COOH + NaCl
So, one mole HCl required to completely convert the one mole C6H5COONa to one mole benzoic acid (C6H5COOH)
So, 0.200 moles of HCl required to completely reacts with 0.200 moles of C6H5COONa.
Volume of HCl = moles of HCl/molarity of HCl = 0.200 mol/6.0 M = 0.200 mol/(6.0 mol/L) = 0.033 L = 33 mL
So, the volume of HCl need to add is 33 mL.