A 0.120 M solution of an enantiomerically pure chiral compound D has an observed
ID: 551638 • Letter: A
Question
A 0.120 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.12 in a 1-dm sample container. The molar mass of the compound is 178.0 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 0.120 M in L, the enantiomer of D? Number deg mL g- dm Number deg (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)? Number Number deg mL g dm deg (e) What is the specific rotation of L, the enantiomer of (f) What is the observed rotation of 100 mL of a solution that contains 0.01 mole of D and 0.005 after the dilution described in part (c)? mole of L? (Assume a 1-dm path length.) Number Number deg mL g dmExplanation / Answer
a) we know that
Specific rotation = (observed rotation)/(concentration)(length of container)
Where concentraiton is in g/mL
given concentration = 0.120 M = 0.120mol/ L = 0.120 X 178 g / L = 21.36g / L = 0.02136g /mL
Therefore
Specific rotation = 0.120/ ( 0.02136 g/mL)(1.0 dm) = +5.62 deg mL/g . dm
(b). We know that if we add same amount of other enantiomer then we will obtain a racemic mixture which is optically inactive and has optical rotation = 0. [Racemic mixture: equimolar mixture of enantiomers)
(c). As we have added same volume of water so the concentration will become half of original concentration (volume is doubled, moles / L will be halved)
New concentration of diluted soluiton = 0.120 / 2 = 0.06 M
Let us convert it into g/mL:
0.06 moles/ L = 0.06 X 178 g /L = 10.68g /L = 10.68 / 1000 g /mL = 0.01068g/mL
Specific rotation = (observed rotation)/(concentration)(length of container)
observed rotation = (specific rotation X (concentration) X (length of container)
observed rotation = +5.62 deg mL/g . dm X 0.01068g/mL X 1.0 dm =0 .0600216 0= this is half of previous optical rotation.
(d) The specific rotation will not affected by dilution, so this will remain the same = +5.62 deg mL/g . dm
(e) The specific rotation of an enantiomer is equal but opposite in sign of specific rotation of other enantiomer. so the specfic rotation of "L" isomer will be = -5.62 0 mL/g . dm
(f) 0.01 of D are taken and 0.005moles of L are taken
So the 0.005 moles of L will cancel the rotation due to 0.005moles of D enantiomer
The roation will be due to only 0.005moles of D.
Let us convert mole /100mL to g/mL
0.005moles / 100mL = 0.005 X 178 g /100mL = 0.89g / 100mL = 0.0089g /mL
observed rotation:
observed rotation = (+5.62 deg mL/g . dm) X 0.0089g /mL X 1 = = +0.050018 degrees
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