A 0.100 M solution of an enantiomerically pure chiral compound D has an observed

Question

A 0.100 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.16 degree in a 1-dm sample container. The molar mass of the compound is 120.0 g/mol. (a) What is the specific rotation of D? (c) What is the observed rotation if the solution of D is diluted with an equal volume of solvent? (e) What is the specific rotation of L, the enantiomer of D, after the dilution described in part (c)? (b) What is the observed rotation if this solution is mixed with an equal volume of a solution that is 0.100 M in L, the enantiomer of D? (d) What is the specific rotation of D after the dilution described in part (c)? (f) 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 path length.)

Explanation / Answer

A) 0.1 moles/L * 120 g/mole * 1L/1000 mL = 0.012 g/mL = 83.33 mL/g +0.16 degrees * 83.33 mL/g dm = +13.33 degrees B) 0 C) 0.08 D) +13.33 degrees E) -13.33 degrees F) 0.01-0.005 = 0.005 moles excess D 0.005 moles/0.1L * 120 g/mole * 1L/1000 mL = 0.006 g/mL = 166.67 mL/g 13.33 = 166.67 * observed rotation observed rotation = 0.08

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