Question

Suppose we dissolve 12.5 g of chiral compound in .2L of solution and place it in a tube that is 20cm long. What is the samples specific rotation of the observed rotation experimentally determined to be 4 degrees?

A) 20 degrees

B) 12.5 degrees

C) .125 degrees

D) 8 degrees

E) 32 degrees

Answer 1

specific rotation $\alpha$    = observed rotation/l*conc (g/ml)

                         l           = 20cm = 2dm

                         conc   = 12.5/200 = 0.0625g/ml

                                     = 4/2*0.0625

                                       = 4/0.125   = 32⁰

             E) 32 degrees