Pure cholesterol has a specific rotation of -32. A sample of cholesterol prepared in the lab has a specific rotation of -8. The enatiomeric excess of the sample of chloresterol is x%. X is :

Answer

To find the enantiomeric excess (ee) of the sample of cholesterol, we can use the formula for enantiomeric excess, which is given by:

\[ \text{Enantiomeric Excess (ee)} = \frac{\text{Observed Specific Rotation}}{\text{Specific Rotation of Pure Enantiomer}} \times 100 \]

### Step 1: Identify the values
- The specific rotation of pure cholesterol is given as \(-32\).
- The observed specific rotation of the lab-prepared cholesterol is given as \(-8\).

### Step 2: Substitute the values into the formula
Using the values we have:

\[ \text{ee} = \frac{-8}{-32} \times 100 \]

### Step 3: Simplify the fraction
The negatives cancel out:

\[ \text{ee} = \frac{8}{32} \times 100 \]

### Step 4: Calculate the fraction
Now, simplify \(\frac{8}{32}\):

\[ \frac{8}{32} = \frac{1}{4} \]

### Step 5: Multiply by 100
Now, multiply by 100 to find the percentage:

\[ \text{ee} = \frac{1}{4} \times 100 = 25 \]

### Conclusion
The enantiomeric excess (ee) of the sample of cholesterol is \(25\%\). Therefore, \(x = 25\).