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

We are given:

- Observed optical rotation of the enantiomeric mixture = $$+12.6°$$

- Specific rotation of the (+) isomer = $$+30°$$

Optical purity (enantiomeric excess) is defined as:

$$\text{Optical purity} = \frac{\text{Observed rotation}}{\text{Specific rotation of pure enantiomer}} \times 100\%$$

Substituting the values:

$$\text{Optical purity} = \frac{12.6}{30} \times 100$$

$$\text{Optical purity} = 0.42 \times 100 = 42\%$$

Therefore, the optical purity is $$\textbf{42}$$ %.