(-)-Cholesterol has a specific rotation of -32o. A mixture of ( )- and (-)-cholesterol was analyzed by polarimetry, and the observed rotation was 14o. What is the percent composition of the ( ) isomer in this mixture

(+)-cholesterol = 71.88%

(-)-cholesterol = 28.12%

Explanation:

Asuming 1 gram of sample is dissolved in 1mL of water and the sample cell was 1dm long.

Enantiomeric excess is defined as the amount of pure enantiomer in a sample. The formula is:

ee = [α]mixture / [α]pure enantiomer.

Replacing:

ee = 14° / 32°×100 = 43.75%

As the sample is 14°, There is an excess of (+)-cholesterol and 56.25% is a 1:1 mixture of enantiomers.

That means percent composition of enantiomers is:

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

it is b that is the answer.