The optical rotation of the `alpha`-form of a pyranose is `+150.7^(@)`, that of the `beta`-form is `+52.8^(@)`. In solution an equilibrium mixture of these anomers has an optical rotation of `+80.2^(@)`. The precentage of the `alpha`-form in equilibrium mixture is :

Options

A. 0.28
B. 0.32
C. 0.68
D. 0.72

Answer

To solve the problem, we need to find the percentage of the alpha-form of pyranose in an equilibrium mixture given its optical rotations. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Known Values**:
- Optical rotation of the alpha-form (α): +150.7°
- Optical rotation of the beta-form (β): +52.8°
- Optical rotation of the equilibrium mixture: +80.2°

2. **Let x be the percentage of the alpha-form**:
- Therefore, the percentage of the beta-form will be (100 - x).

3. **Set up the equation for optical rotation**:
The optical rotation of the mixture can be expressed as:
\[ \frac{x \cdot (\text{optical rotation of } \alpha) + (100 - x) \cdot (\text{optical rotation of } \beta)}{100} = \text{optical rotation of mixture} \]
Substituting the known values:
\[ \frac{x \cdot 150.7 + (100 - x) \cdot 52.8}{100} = 80.2 \]

4. **Multiply through by 100 to eliminate the denominator**:
\[ x \cdot 150.7 + (100 - x) \cdot 52.8 = 8020 \]

5. **Distribute the terms**:
\[ 150.7x + 5280 - 52.8x = 8020 \]

6. **Combine like terms**:
\[ (150.7 - 52.8)x + 5280 = 8020 \]
\[ 97.9x + 5280 = 8020 \]

7. **Isolate x**:
\[ 97.9x = 8020 - 5280 \]
\[ 97.9x = 2740 \]

8. **Solve for x**:
\[ x = \frac{2740}{97.9} \approx 27.94 \]

9. **Convert to percentage**:
Since x is the percentage of the alpha-form:
\[ \text{Percentage of alpha-form} \approx 27.94\% \]

### Final Answer:
The percentage of the alpha-form in the equilibrium mixture is approximately **28%**.