In: Chemistry
(-)-Mandelic acid has a specific rotation of -158°. What would be the specific rotation of a solution which contains 40% (-)-mandelic acid and 60% (+)-mandelic acid?
+95° |
-32° |
-63° |
+63° |
+32° |
Given the % of (-)-mandelic acid is 40%
its specific rotation = -158o
Given the % of (+)-mandelic acid is 60%
its specific rotation = +158o
The specific rotation of the solution = (% of - rotation x its percentage)+(% of + rotation x its percentage)
= [(40/100)x(-158) ] + [(60/100) x (+158)]
= +32 o