A sugar exist in two isomeric form `alpha-D` & `beta-D` with specific rotation of `110^(@)` and `30^(@)` respectively. When `alpha-D` form of the suar is dissolved in water exhibits mutarotation till the equilibrium value of `90^(@)` is attained. The equilibrium constant of mutarotation of `beta-D` form [Assuming only `alpha` and `beta` form in solution] is equal to:

Explanation

To solve the problem, we need to determine the equilibrium constant for the mutarotation of the sugar from its alpha-D form to its beta-D form. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Specific rotation of α-D: \( [\alpha]_{\alpha} = 110^\circ \)
- Specific rotation of β-D: \( [\alpha]_{\beta} = 30^\circ \)
- Equilibrium specific rotation after mutarotation: \( [\alpha]_{eq} = 90^\circ \)

2. **Setting Up the Equilibrium**:
- Let \( x \) be the fraction of the α-D form at equilibrium.
- Then, the fraction of the β-D form at equilibrium will be \( 1 - x \).

3. **Using the Specific Rotation Formula**:
- The specific rotation at equilibrium can be expressed as:
\[ [\alpha]_{eq} = x \cdot [\alpha]_{\alpha} + (1 - x) \cdot [\alpha]_{\beta} \]
- Substituting the known values:
\[ 90 = x \cdot 110 + (1 - x) \cdot 30 \]

4. **Solving for \( x \)**:
- Expanding the equation:
\[ 90 = 110x + 30 - 30x \]
- Simplifying:
\[ 90 = 80x + 30 \]
- Rearranging gives:
\[ 80x = 90 - 30 = 60 \]
- Therefore:
\[ x = \frac{60}{80} = \frac{3}{4} \]

5. **Finding the Equilibrium Constant \( K_c \)**:
- The equilibrium constant for the mutarotation can be defined as:
\[ K_c = \frac{[\beta]}{[\alpha]} = \frac{1 - x}{x} \]
- Substituting \( x = \frac{3}{4} \):
\[ K_c = \frac{1 - \frac{3}{4}}{\frac{3}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

6. **Final Answer**:
- The equilibrium constant \( K_c \) for the mutarotation of the β-D form is:
\[ K_c = \frac{1}{3} \approx 0.33 \]