Natural boron is a mixture of two isotopes `""_(5)B^(10)` and `""_(5)B^(11)` of masses 10.003 amu and 11.009 amu, respectively. Calculate the atomic mass of natural boron if the relative abundance of `""_(5)B^(10)` and `""_(5)B^(11)` is 19.97% and 80.03%, respectively.

Answer

To calculate the atomic mass of natural boron, we will use the weighted average formula based on the relative abundances of its isotopes. Here are the steps to find the solution:

### Step 1: Identify the isotopes and their masses
We have two isotopes of boron:
- Isotope \( ^{10}B \) with a mass of \( 10.003 \, \text{amu} \)
- Isotope \( ^{11}B \) with a mass of \( 11.009 \, \text{amu} \)

### Step 2: Identify the relative abundances
The relative abundances of the isotopes are given as:
- \( ^{10}B \): \( 19.97\% \)
- \( ^{11}B \): \( 80.03\% \)

### Step 3: Convert percentages to fractions
To use these percentages in calculations, convert them to fractions:
- \( \text{Fraction of } ^{10}B = \frac{19.97}{100} = 0.1997 \)
- \( \text{Fraction of } ^{11}B = \frac{80.03}{100} = 0.8003 \)

### Step 4: Calculate the weighted average
The atomic mass of natural boron can be calculated using the formula:
\[ \text{Atomic mass of natural boron} = ( \text{Fraction of } ^{10}B \times \text{Mass of } ^{10}B ) + ( \text{Fraction of } ^{11}B \times \text{Mass of } ^{11}B ) \]
Substituting the values:
\[ \text{Atomic mass of natural boron} = (0.1997 \times 10.003) + (0.8003 \times 11.009) \]

### Step 5: Perform the calculations
Calculating each term:
1. \( 0.1997 \times 10.003 = 1.997 \)
2. \( 0.8003 \times 11.009 = 8.803 \)

Now, add these two results:
\[ \text{Atomic mass of natural boron} = 1.997 + 8.803 = 10.800 \, \text{amu} \]

### Final Result
The atomic mass of natural boron is approximately \( 10.800 \, \text{amu} \).

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