Naturally occurring boron consists of two isotopes whese atomic weights are `10.01` and `11.01`. The atomic weight of natural boron is `10.81`. Calculate the percentage of each isotope in natural boron.

Explanation

To solve the problem of finding the percentage of each isotope in naturally occurring boron, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Isotopes and Their Weights**:
- Isotope 1 (Boron-10): Atomic weight = 10.01
- Isotope 2 (Boron-11): Atomic weight = 11.01
- Average atomic weight of natural boron = 10.81

2. **Define the Variables**:
- Let the percentage (relative abundance) of Isotope 1 (Boron-10) be \( \alpha \).
- Therefore, the percentage of Isotope 2 (Boron-11) will be \( 100 - \alpha \).

3. **Set Up the Equation for Average Atomic Weight**:
The average atomic weight can be expressed as:
\[ \text{Average Atomic Weight} = \frac{(10.01 \times \alpha) + (11.01 \times (100 - \alpha))}{100} \]
Given that the average atomic weight is 10.81, we can set up the equation:
\[ 10.81 = \frac{(10.01 \times \alpha) + (11.01 \times (100 - \alpha))}{100} \]

4. **Clear the Denominator**:
Multiply both sides of the equation by 100 to eliminate the fraction:
\[ 10.81 \times 100 = 10.01 \times \alpha + 11.01 \times (100 - \alpha) \]
This simplifies to:
\[ 1081 = 10.01 \alpha + 1101 - 11.01 \alpha \]

5. **Combine Like Terms**:
Rearranging the equation gives:
\[ 1081 = 1101 - 0.99 \alpha \]
Now, isolate \( \alpha \):
\[ 0.99 \alpha = 1101 - 1081 \]
\[ 0.99 \alpha = 20 \]

6. **Solve for \( \alpha \)**:
Divide both sides by 0.99:
\[ \alpha = \frac{20}{0.99} \approx 20.20 \]

7. **Calculate the Percentage of Isotope 2**:
Since \( \alpha \) represents the percentage of Isotope 1:
\[ \text{Percentage of Isotope 2} = 100 - \alpha \approx 100 - 20.20 \approx 79.80 \]

8. **Final Results**:
- Percentage of Isotope 1 (Boron-10) ≈ 20.20%
- Percentage of Isotope 2 (Boron-11) ≈ 79.80%

### Summary:
- The percentage of Boron-10 is approximately **20.20%**.
- The percentage of Boron-11 is approximately **79.80%**.