Answer:
the enantiomeric excess of the mixture is 40%
Explanation:
The computation of the enantiomeric excess of the mixture is shown below:
As we know that
[tex]= |\frac{R - S}{R + S} |\times 100\\\\= |\frac{70 - 30}{70 + 30}| \times 100\\\\= 40\%[/tex]
Hence, the enantiomeric excess of the mixture is 40%
