Question

Optical activity of an enantiomeric mixture is $+12.6^{\circ }$ and the specific rotation of $(+)$ isomer is $+30^{\circ }$. The optical purity is ______$\%$

Answer 1

Answer: 42

$\%$ optical purity
$=\frac{\text{ observed rotation of mixture }\times 100}{\text{ rotation of pure enantiomer }}$
$=\frac{+12.6^{\circ }}{+30^{\circ }}\times 100=42$