A tube of sugar solution 20 cm long is placed between crossed nicols and illuminated with light of wavelength `6xx10^(-5)` cm If the optical rotation produced is `13^(@)` and the specific rotation is `65^(@)` determine the strength of the solution

Options

A. `0.1g//cc`
B. `0.2g//cc`
C. `0.9g//cc`
D. `1.0g//cc`

Explanation

To solve the problem, we will use the formula for specific rotation, which relates the optical rotation, the length of the tube, and the concentration of the solution.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the tube (L) = 20 cm
- Optical rotation (θ) = 13°
- Specific rotation (α) = 65°

2. **Convert Length to Decimeters:**
- Since the specific rotation formula requires the length in decimeters, we convert 20 cm to decimeters.
\[ L = \frac{20 \text{ cm}}{10} = 2 \text{ dm} \]

3. **Use the Formula for Specific Rotation:**
The specific rotation is given by the formula:
\[ \alpha = \frac{\theta}{L \cdot C} \]
Rearranging this formula to find the concentration (C):
\[ C = \frac{\theta}{L \cdot \alpha} \]

4. **Substitute the Values into the Formula:**
Now, substituting the values we have:
\[ C = \frac{13°}{2 \text{ dm} \cdot 65°} \]

5. **Calculate the Concentration:**
- First, calculate the denominator:
\[ 2 \cdot 65 = 130 \]
- Now, divide the optical rotation by this value:
\[ C = \frac{13}{130} = \frac{1}{10} = 0.1 \text{ g/L} \]

6. **Conclusion:**
The strength of the sugar solution is **0.1 g/L**.