Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with a half life of `3.33` hrs at `25^(@)C` What fraction of sample of sucrose remains after `9.00` hrs

Options

A. `0.333`
B. `0.666`
C. `0.153`
D. `0.250`

Explanation

To solve the problem of how much sucrose remains after 9 hours given its half-life and that it follows first-order kinetics, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Half-life (t₁/₂) = 3.33 hours
- Total time (t) = 9.00 hours

2. **Calculate the Rate Constant (k)**:
For a first-order reaction, the rate constant (k) can be calculated using the half-life formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the given half-life:
\[ k = \frac{0.693}{3.33} \approx 0.208 \, \text{hr}^{-1} \]

3. **Determine the Number of Half-Lives in 9 Hours**:
To find out how many half-lives fit into 9 hours:
\[ \text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{9.00}{3.33} \approx 2.70 \]

4. **Calculate the Fraction Remaining**:
The fraction of the original substance remaining after n half-lives can be calculated using the formula:
\[ \text{Fraction remaining} = \left(\frac{1}{2}\right)^n \]
Substituting the number of half-lives:
\[ \text{Fraction remaining} = \left(\frac{1}{2}\right)^{2.70} \approx 0.153 \]

5. **Conclusion**:
The fraction of the sucrose sample that remains after 9 hours is approximately **0.153**.