0.150 M solution olan enantiomerically pure chiral compound D has an observed ro
ID: 866934 • Letter: 0
Question
0.150 M solution olan enantiomerically pure chiral compound D has an observed rotation of +0.14 Degree in a 1-dm sample container. The molar mass of the compound is 152.0 g/mol. What is the specific rotation ofD? What is the observed rotation If this solution IS mixed with an equal volume of a solution that is 0.150 M in L.the enanliomer of D? What is the observed rotation if the solution of D is diluled with an equal volume of solvent? What is the speoific rotation of D after the dilution described in part (c)? What is the specific rotation of L. Ihe enantlomer of D. after the dilution described in part (c)? What Is the observed rotation of 100 mL of a solution that contains 0.01 mole of D and 0.005 mole of L? (Assume a 1- dm palh length.)Explanation / Answer
Specific rotation = observed rotation / conc. (g/mL) x path length (dm)
(a) We have +0.14 observed rotation
C = 0.15 M = 0.15 M/L = 0.15 x 152 x 0.001 = 0.0228 g/ml
We have, path length = 1 dm
Thus, specific rotation = +0.14 / (0.0228 x 1) = +6.14
(b) The observed rotation will not change and it will be = +0.14
(c) The observed rotation If diluted will be = +0.14/2 = +0.07
(d) Secific rotation after dilution will be = +0.07 / (0.000228 x 1) = +307
(e) The specific rotation of L will be = -307
(f) The concentration of the solution will be 0.005 M of D, 0.005 M of L racemises 0.005 M of D.
Thus, observed rotation will be = 4.7
c = 0.0001 M = 0.0001 moles L-1 = (0.0001) (853.93 g mole-1) (0.001 L) = 0.000085 g ml-1
Assume l = 1.0 dm (standard polarimeter tube)
Solving for ?: ? = [?]lTlc = (-49o) (1.0 dm) (0.000085 g ml-1) = -0.0042o.
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