Question

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?

Answer 1

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.