A sample of optically active alcohol has a specific rotation, `[alpha]_(D)^(25)`, equal to `+1.151^(@)`. Specific rotation for pure form `[alpha]_(D)^(25)` is `+5.756^(@)`.
What is the percentage enantiomeric excess of the sample?

Explanation

To find the percentage enantiomeric excess of the sample, we can use the following formula:

\[ \text{Percentage Enantiomeric Excess} = \left( \frac{\text{Observed Rotation}}{\text{Specific Rotation of Pure Enantiomers}} \right) \times 100 \]

### Step-by-step Solution:

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