The natural boron of atomic weight 10.81 is found to have two isotopes `.^10B` and `.^11B` .The ratio of abundance of isotopes of natural boron should be

Options

A. `11:10`
B. `81:19`
C. `10:11`
D. `19:81`

Explanation

To find the ratio of the abundance of the isotopes of natural boron, we can follow these steps:

### Step 1: Define the variables
Let:
- \( x \) = percentage abundance of \( ^{10}B \)
- \( 100 - x \) = percentage abundance of \( ^{11}B \)

### Step 2: Set up the equation for average atomic weight
The average atomic weight of natural boron is given as 10.81. We can express this as:
\[ \text{Average Atomic Weight} = \frac{(10 \times x) + (11 \times (100 - x))}{100} \]
Substituting the known average atomic weight:
\[ 10.81 = \frac{(10x) + (11(100 - x))}{100} \]

### Step 3: Clear the denominator
Multiply both sides by 100 to eliminate the fraction:
\[ 1081 = 10x + 1100 - 11x \]

### Step 4: Simplify the equation
Combine like terms:
\[ 1081 = 1100 - x \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x = 1100 - 1081 \]
\[ x = 19 \]

### Step 6: Find the abundance of \( ^{11}B \)
Now, we can find the abundance of \( ^{11}B \):
\[ 100 - x = 100 - 19 = 81 \]

### Step 7: Write the ratio
The ratio of the abundance of \( ^{10}B \) to \( ^{11}B \) is:
\[ \text{Ratio} = 19 : 81 \]

### Conclusion
Thus, the ratio of the abundance of isotopes of natural boron is \( 19 : 81 \).

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