Let `(+)` tartaric acid has a specific rotation of`+11.1`, Calculate the specific rotation of a mixture of `68%(+)` tartaric acid and `32%(-)` tartaric acid.

Explanation

To solve the problem of calculating the specific rotation of a mixture of 68% (+) tartaric acid and 32% (-) tartaric acid, we will follow these steps:

### Step 1: Identify the specific rotations
We know that:
- The specific rotation of (+) tartaric acid is +11.1.
- The specific rotation of (-) tartaric acid is -11.1 (since it is the enantiomer).

### Step 2: Calculate the enantiomeric excess (EE)
The enantiomeric excess can be calculated using the formula:
\[ \text{EE} = \frac{D - L}{D + L} \times 100 \]
Where:
- \(D\) = percentage of (+) tartaric acid = 68%
- \(L\) = percentage of (-) tartaric acid = 32%

Substituting the values:
\[ \text{EE} = \frac{68 - 32}{68 + 32} \times 100 = \frac{36}{100} \times 100 = 36\% \]

### Step 3: Calculate the optical rotation of the mixture
The optical rotation of the mixture can be calculated using the formula:
\[ \text{Optical Rotation} = \text{EE} \times \text{Specific Rotation of the excess enantiomer} \]
Since the enantiomeric excess is 36% and the specific rotation of the excess enantiomer (+) tartaric acid is +11.1, we have:
\[ \text{Optical Rotation} = 36\% \times 11.1 = 4 \]

### Step 4: Calculate the specific rotation of the mixture
The specific rotation of the mixture can be calculated using the formula:
\[ [\alpha] = \frac{\text{Optical Rotation}}{\text{Total Concentration}} \]
In this case, the total concentration is 100% (68% + 32%). Thus:
\[ [\alpha] = \frac{4}{100} = 0.04 \]

### Final Answer
The specific rotation of the mixture of 68% (+) tartaric acid and 32% (-) tartaric acid is **4**.

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