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 +0.14 degrees in a 1-dm sample container. The molar mass of the compounds is 140.0g/mol

A) What is the specific rotation of D?

B) What is the observed rotation if this solution is mixed with an equal volume that is 0.100 M in L, L is 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 diltuion descibed 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 / (conc,g/ml) (length of sample tube, decimeters)

so, given observed rotaion = +0.14

conc = 0.1M= no of moles / volume in L

so mass taken/molar mass x volume in L = 0.1M

mass taken per volume in L = 0.1 x 140 g/mol = 14g/L

conc ( g/mL ) = 0.014g/mL

length = 1dm

So specific rotation = 0.14 / (0.014x1) = 10 degrees

B. when both enentiomers are in equal amounts of equal concentrations, external compensation take place and Observed rotation is ZERO.

C. Please provide the nature of solvent as rotation depends on the chiral nature of the solvent. ( so cannot answer C-E)

F.Observed Rotation = specific Rotation xconc,g/ml) x(length of sample tube, decimeters)

As specific rotation is a same for a substance , it is 10 degrees.

conc = 0.01mol in 100mL = 1.4g in 100mL = 0.014g/mL

length = 1dm

So, observed rotaion = 10 x 0.014 x 1 = + 0.14 degrees.

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