If solution of a compound (`30g//100ml` of solution ) has measured rotation of `+15^(@)` in a `2 dm` long sample tube, the specific rotation of this compound is

Options

A. `+50^(@)`
B. `+7.5^(@)`
C. `+15^(@)`
D. `+25^(@)`

Explanation

To find the specific rotation of the compound, we can use the formula for specific rotation, which is given by:

\[ [\alpha] = \frac{\alpha}{c \cdot l} \]

where:
- \([\alpha]\) is the specific rotation,
- \(\alpha\) is the observed rotation (in degrees),
- \(c\) is the concentration of the solution (in g/mL),
- \(l\) is the path length of the sample tube (in dm).

### Step-by-Step Solution:

1. **Identify the given values:**
- Observed rotation (\(\alpha\)) = +15 degrees
- Mass of the compound = 30 g
- Volume of the solution = 100 mL
- Length of the sample tube (\(l\)) = 2 dm

2. **Calculate the concentration (\(c\)):**
\[ c = \frac{\text{mass of solute}}{\text{volume of solution}} = \frac{30 \text{ g}}{100 \text{ mL}} = 0.3 \text{ g/mL} \]

3. **Substitute the values into the specific rotation formula:**
\[ [\alpha] = \frac{15 \text{ degrees}}{0.3 \text{ g/mL} \times 2 \text{ dm}} \]

4. **Calculate the denominator:**
\[ 0.3 \text{ g/mL} \times 2 \text{ dm} = 0.6 \text{ g \cdot dm/mL} \]

5. **Now substitute this back into the specific rotation formula:**
\[ [\alpha] = \frac{15}{0.6} \]

6. **Perform the division:**
\[ [\alpha] = 25 \text{ degrees} \]

7. **Final Result:**
The specific rotation of the compound is \([\alpha] = +25^\circ\).