A 0.100 M solution of an enantiomercially pure chiral compound D has an observed

Question

A 0.100 M solution of an enantiomercially pure chiral compound D has an observed rotation of +.26 degree in a 1-dm sample container. The molar mass of the compount is 143.0 g/mol.

a. What is the specific rotation of D? (deg mL/ g.dm)

b. What is the observed rotation if this solution is mixed with an equal volume of a solution that is .100 M in L, the enantiomer of D? (deg)

c. What is the observed rotation if the solution of D is diluted with an equal volume of solvent? (degree)

d. What is the specific rotation of D after the dilution described in part c? (deg mL/g.dm)

e. What is the specific rotation of L, the enantiomer of D, after the dilution described in part c? (deg mL/g.dm)

f. What is the observed rotation of 100mL of a solution that contains .01 mole of D and .005 mole of L? (Assume a 1-dm path length.) (degree)

Explanation / Answer

a)

concentration = 0.100 x 143 = 14.3 g / L = 0.0143 g/mL

observed rotation = + 0.26 o

specific rotation = observed rotation / l x concnetration

                             = 0.26 / 1 x 0.0143

specific rotation = 18.2 deg .mL / g.dm

b)

equial volume and equal concnetrations of solutions mixed , racemic mixure formed.

observed rotation = 0 deg

c)

volume doubled , then concentration is halved.

c = 0.0143 / 2 = 0.00715

observed rotation = 18.2 x 0.00715 x 1

observed rotation = 0.13 deg

d)

specific rotation is uneffected by dilution

specific rotation = 18.2

e)

specific rotation of L = - 18.1

f)

moles= 0.01 - 0.005 = 0.005

concentration = 0.005 / 0.1 = 0.05 M

                       = 0.05 x 143 / 1000

                       = 0.00715

observed rotation = 0.00715 x 18.2 x 1

observed rotation = 0.13

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