A 0.140 M solution of an enantiomerically pure chiral compound D has an observed
ID: 534776 • Letter: A
Question
A 0.140 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.20 degree in a 1-dm sample container. The molar mass of the compound is 123.0 g/mol. (a) What is the specific rotation of D? (b) What is the observed rotation if this solution is mixed with an equal volume of a solution that is 0.140 M in L, the enantiomer of D? (c) What is the observed rotation if the solution of D is diluted with an equal volume of solvent? (d) What is the specific rotation of D after the dilution described in part (c)? (e) What is the specific rotation of L, the enantiomer 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).
Specific rotation = observed rotation/(concentration)(length of container)
concentration = 0.140 mol / 1000 mL x 123 = 0.01722 g / mL
Specific rotation = 0.20 /(0.02235)(1.0 )
Specific rotation = + 11.61 degrees mL/g dm
(b).
If we add the same concentration of D's enantiomer, L, we will have a racemic mixture.
The observed rotation then will be = 0.
(c).
concentration = 0.140 / 2 = 0.070 M:
(0.07 mol/1000 mL)(123.0 g/mol) = 8.61 x 10^-3 g/mL
observed rotation = (11.61 )(8.61 x 10^-3 )(1.0 dm)
observed rotation = + 0.10 deg
(d)
The specific rotation will not be affected by the dilution
,so
specific rotation = + 11.61 degrees mL/g dm
(e)
The specific rotation of L will simply be the negative of D's specific rotation
L's specific rotation will be = - 11.61 degrees mL/g dm.
f)
observed rotation = 0.0714
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