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postnew [5]
3 years ago
5

Consider a solution that contains 60.0% r isomer and 40.0% s isomer. if the observed specific rotation of the mixture is –43.0°,

what is the specific rotation of the pure r isomer?
Chemistry
1 answer:
pychu [463]3 years ago
5 0
The solution for this problem is:
Let x denote the specific rotation, R; andLet y denote the specific rotation, S = -x 
Solution:60 x - 40 x/100 = - 43
20x = - 4300Divide both sides by 20The answer is:x = - 215 is the specific rotation of the pure r isomer.
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