If a pure R isomer has a specific rotation of -137.0 degrees and a sample contains...

If a pure R isomer has a specific rotation of -137.0 degrees and a sample contains 84.0% of the R siomer and 16.0% of it anantiomer, what is the observed specific rotation of the mixture?

Expert Solution

obsereved specific rotation

if R = -137o

S = +1 37

observed specific rotation of the mixture = R specific rotation + S -specific rotation

= 84 /100 * -137 + 16 / 100 * 137 = -115.08 + 21.92

= -93.16

observed specific rotation of the mixture = -93.16 ^o

queen_honey_blossom answered 3 weeks ago

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