Question

A solution prepared by mixing 20 ml of a 0.10 M solution of the Renantiomer of a compound and 60 mL of a 0.10 M solution of the Senartiomer was found to have an observed specific rotation of 3.4. (Hint: ml x M millimolo, abbreviated as mmol) Part A What is the specific rotation of Senantiomer? Express your answer using one decimal place. AED RO? Submit Request Answer

Answer 1

Volume of R enantioner = 20 mL concentration af Renantiomney = 0:10M volum of s enantiomer = Gome concentration of s- enantiomer = 0.10M on mixing of solution, volume of solution gets incuared concentration of added solutions also get changed a Nero a are concentration calculated as Final concentration ass- enantiomer = concox volume Total volumea neue = [s] final = 6 milimal 80 ML 0.10 M X bomL 20mL +60mL 0.075 M I [s] final = Similarly [Rifinal = OlOM x 20mL 20mL + 60ml 0 2 milimol 80m2 TRI final = 0.025M specific rotation is given by [+] = x 111111 wheru a c= 1= = obsewed specific rotation = +3.4° concentration of sample lenges of tube - to dm

specific-rotation of s- enantiomer is Is- enantion = +304 deg Idmx 0-075 molt W = is [x] so enantion = 45030 h specific rotation R- enantiomer [x] = - +3040 Idm x 0.025m Tar = 3.4 0.025 [ <DR = 1361