Question

(-)-Mandelic acid has a specific rotation of -158°. What would be the specific rotation of a solution which contains 40% (-)-mandelic acid and 60% (+)-mandelic acid?

+95°

-32°

-63°

+63°

+32°

Answer 1

Given the % of (-)-mandelic acid is 40%

its specific rotation = -158^o

Given the % of (+)-mandelic acid is 60%

its specific rotation = +158^o

The specific rotation of the solution = (% of - rotation x its percentage)+(% of + rotation x its percentage)

                                                   = [(40/100)x(-158) ] + [(60/100) x (+158)]

                                                   = +32 ^o