If a pure R isomer has a specific rotation of –146.0°, and a sample contains 80.0%...

If a pure R isomer has a specific rotation of –146.0°, and a sample contains 80.0% of the R isomer and 20.0% of its enantiomer, what is the observed specific rotation of the mixture?

Expert Solution

observed specific rotation of the mixture = (specific rotation of R isomer X percent of R isomer) +(specific rotation of R isomer X percent of enantiomer)

observed specific rotation of the mixture = (- 146 X 80/100 + 146 X 20/100)

= - 116.8 + 29.2

= - 87.6 0°

queen_honey_blossom answered 6 months ago

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