An organic compound P exists in two enantiomeric forms, which have specific optical rotation values `[alpha] = +-100^(@)`. The optical rotation of a mixture of these two enantiomers is `-50^(@)`. Calculate the percentage of that enantiomer which is in lower concentration in the mixture.

Explanation

To solve the problem, we need to determine the percentage of the enantiomer that is present in lower concentration in a mixture of two enantiomers of compound P. The specific optical rotations of the enantiomers are given as +100° and -100°, and the observed optical rotation of the mixture is -50°.

### Step-by-Step Solution:

1. **Understanding the Optical Rotation Values**:
- The specific rotation of the dextrorotatory enantiomer (let's call it D) is +100°.
- The specific rotation of the levorotatory enantiomer (let's call it L) is -100°.
- The observed rotation of the mixture is -50°.

2. **Using the Enantiomeric Excess Formula**:
- The enantiomeric excess (EE) can be calculated using the formula:
\[ EE = \frac{\text{Observed Rotation}}{\text{Specific Rotation of Pure Enantiomer}} \times 100 \]
- Since the observed rotation is -50° and we will use the absolute value of the specific rotation of the pure enantiomer (which is 100°), we can substitute:
\[ EE = \frac{-50}{100} \times 100 = -50\% \]

3. **Calculating the Concentrations**:
- The enantiomeric excess indicates that the mixture is not racemic. The total percentage of the two enantiomers must add up to 100%.
- Let \( x \) be the percentage of the dextrorotatory isomer (D), and \( y \) be the percentage of the levorotatory isomer (L).
- We know:
\[ x + y = 100\% \]
- The enantiomeric excess can also be expressed as:
\[ EE = \frac{x - y}{x + y} \times 100 \]
- Substituting \( x + y = 100\% \):
\[ EE = \frac{x - (100 - x)}{100} \times 100 = \frac{2x - 100}{100} \times 100 = 2x - 100 \]
- Setting this equal to -50%:
\[ 2x - 100 = -50 \]
- Solving for \( x \):
\[ 2x = 50 \implies x = 25\% \]

4. **Finding the Percentage of the Lower Concentration Enantiomer**:
- Since \( x \) (the percentage of D) is 25%, the percentage of L (the levorotatory isomer) is:
\[ y = 100 - x = 100 - 25 = 75\% \]
- Therefore, the percentage of the enantiomer in lower concentration (D) is **25%**.

### Final Answer:
The percentage of the enantiomer which is in lower concentration in the mixture is **25%**.