In a mixture, two enantiomers are found to be present in $85$ % and $15$ % respectively. The enantiomeric excess $(e, e^{'})$ is :

85

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15

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70

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60

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Solution

The correct option is C $70$ %
In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e) is $85 - 15 = 70$ %
Note: Enantiomeric excess is the difference between the percent of two enantiomers.

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In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e′) is :

The correct option is C 70%In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e) is 85−15=70 %Note: Enantiomeric excess is the difference between the percent of two enantiomers.

In a mixture, two enantiomers are found to be present in $85$ % and $15$ % respectively. The enantiomeric excess $(e, e^{'})$ is :

The correct option is C $70$ %
In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (e,e) is $85 - 15 = 70$ %
Note: Enantiomeric excess is the difference between the percent of two enantiomers.