Question

A 0.140 M solution of an enantiomerically pure chiral compound D has an observed rotation of 0.18° in a 1-dm sample container. The molar mass of the compound is 150.0 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 0.140 M in L, 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 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.)

Answer 1

a)

concentration = 0.140 x 150 = 21 g / L = 0.021 g/mL

observed rotation = + 0.18 o

specific rotation = observed rotation / l x concnetration

                             = 0.18 / 1 x 0.021

specific rotation = 8.57 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 = 0.021 / 2 = 0.0105

observed rotation = 8.57 x 0.0105 x 1

observed rotation = 0.090 deg

d)

specific rotation is uneffected by dilution

specific rotation = 8.57

e)

specific rotation of L = - 8.57

f)

moles= 0.01 - 0.005 = 0.005

concentration = 0.005 / 0.1 = 0.05 M

                       = 0.05 x 150 / 1000

                       = 0.0075

observed rotation = 0.0075 x 8.57 x 1

observed rotation = 0.064