(-)-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:

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

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