You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution

ee = 96%

Explanation:

Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the  differences of amounts of enantiomers and the total amunt. that is:

ee = |A-B|/ A+B * 100

ee = |98%-2%| / 98+2 * 100

the answer to your question is c

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