A 0.110 M solution of an enantiomerically pure chiral compound D has an observed

Question

A 0.110 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.26 in a 1-dm sample container. The molar mass of the compound is 150.0 g/mol.

Map o Organic Chemistry Marc Loudon A 0.110 M solution of an enantiomerically pure chiral 1-dm sample container. The molar mass of the D has an observed rotation of +0.26 in a is 150.0 g/mol (a) What is the specific rotation of D? b) What is the observed rotation if this solutionn is mixed with an equal volume of a solution that is 0.110 M in L, the enantiomer of D? deg mL g dm Number deg (c) What is the observed rotation if the solution D is diluted with an equal volume of solvent? (d) What is the specific rotation of after the dilution described in part click to edit Number Number deg mL . 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 Number deg mL : dm deg Previous Give Up & View Solution D Check Answer NextExit Hint

Explanation / Answer

a.So I understand that I'm using the formula [specific rotation = observed rotation / (c*l)]. So here is how I tried to find c:

.110 M = mol/L
150 g/mol

.11*150 =15.0 g/L* 1 L/ 1000 mL = .0150

So the equation would me +0.26/(.0150)(1)...
...making the specific rotation 17.3333

b. Does this mean I add an additional .110 M to my equation?
Sort of like:

17.3333 = observed rotation/ 2(.0150)

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