A 0.150 M solution of an enantiomerically pure chiral compound D has an ob se dc
ID: 543300 • Letter: A
Question
A 0.150 M solution of an enantiomerically pure chiral compound D has an ob se dc ed rotation of +0.28 degrees in a 1-dm sample container. The molar mass of the compound is 165.0 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 0.150 M in L, the enantiomer in 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.)
HE 201-Fall17 -CLIZBE/GONG Activities and Due Dates Ch 6 O/18/2017 11:55 PM @ 60/10010/18/2017 03:17 PM Print @Calculator-l Periodic Table Grad estion 3 of 20 Map Organic Chemistry Roberts & Company Publishers presented by Sapling Leaming A 0.150 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.28 in a 1-dm sample container. The molar mass of the compound is 165.0 g/mol (a) What is the specific rotation of D? (b) What is the observed rotation if this solution mixed with an equal volume of a solution that is 0.150 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 (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) e) What is the specific rotation of L, the enantiomer of after the dilution described in part (c)? Number deg ml g dm deg O Pritvious Check Answer Next Exi Hint loExplanation / Answer
SOLUTION:
(a). Specific rotation = observed rotation/(concentration)(length of container)
Concentration of the sample is expressed in g/mL, so we must convert M (moles/L) into g/mL, keeping in mind that 1 L = 1000mL:
(0.150 mol/1000mL)(165.0 g/mol) = 0.02475 g/mL
Specific rotation = 0.28 degrees/(0.02475 g/mL)(1.0 dm) = +11.31 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. (Remember that enantiomers rotate polarized light in equal, but opposite, directions).
(c). Since our molarity is 0.150 M or 0.150 moles/ liter of solution, let's assume we start with 1 L. If we add another liter to the solution, we will have 2 liters. The molarity of the sample will be halved, or 0.075 M. Convert this into g/mL:
(0.075 mol/1000 mL)(165.0 g/mol) = 0.01237 g/mL
observed rotation = (specific rotation)(concentration)(length of container)
observed rotation = (11.31 degrees mL/g dm)(0.01237 g/mL)(1.0 dm) = 0.1399degrees = +0.14 degrees, half of our original observed rotation of +0.28.
(d) The specific rotation will not be affected by the dilution described in (c), since it is independent of concentration. It will remain +11.31 degrees mL/g dm
(e) The specific rotation of L will simply be the negative of D's specific rotation, since enantiomers by definition have specific rotations that are equal in magnitude, but opposite in sign.Therefore L's specific rotation will be -11.31 degrees mL/g dm.
(f) The 0.005 moles of L will cancel out the rotation of 0.005 moles of D. It will be as if we had a 100 mL solution of ONLY 0.005 moles of D:
(0.005 moles/100 mL)(165.0 g/mole) =0.00825 g/mL
Now let's find the observed rotation:
observed rotation = (+11.31 degrees mL/g dm)(0.00825 g/mL)(1 dm) = +0.093308 degrees, or +0.1.
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