Indium (atomic mass `=114.82`) has two naturally occurring isotopes, the predominant one from has isotopic mass `114.9041` and abundance of `95.72%`. Which of the following isotopic mass is the most likely for the other isotope ?

Options

A. 112.94
B. 115.9
C. 113.9
D. 114.9

Answer

To solve the problem, we need to find the isotopic mass of the second isotope of indium given the average atomic mass and the isotopic mass and abundance of the first isotope. Let's break it down step by step.

### Step 1: Understand the given data
- Average atomic mass of indium = 114.82
- Isotopic mass of the predominant isotope (Isotope 1) = 114.9041
- Abundance of Isotope 1 = 95.72%
- Abundance of Isotope 2 = 100% - 95.72% = 4.28%

### Step 2: Set up the equation for average atomic mass
The average atomic mass can be calculated using the formula:
\[ \text{Average Atomic Mass} = \left(\text{Isotopic Mass}_1 \times \frac{\text{Abundance}_1}{100}\right) + \left(\text{Isotopic Mass}_2 \times \frac{\text{Abundance}_2}{100}\right) \]
Substituting the known values:
\[ 114.82 = \left(114.9041 \times \frac{95.72}{100}\right) + \left(x \times \frac{4.28}{100}\right) \]

### Step 3: Calculate the contribution of Isotope 1
Calculate the contribution of Isotope 1 to the average atomic mass:
\[ \text{Contribution of Isotope 1} = 114.9041 \times 0.9572 = 109.986 \]

### Step 4: Substitute the contribution into the equation
Now substitute this value back into the equation:
\[ 114.82 = 109.986 + \left(x \times 0.0428\right) \]

### Step 5: Isolate the variable \(x\)
Rearranging the equation to isolate \(x\):
\[ x \times 0.0428 = 114.82 - 109.986 \]
\[ x \times 0.0428 = 4.834 \]

### Step 6: Solve for \(x\)
Now, divide both sides by 0.0428 to find \(x\):
\[ x = \frac{4.834}{0.0428} \approx 112.94 \]

### Conclusion
The isotopic mass of the other isotope of indium is approximately **112.94**.

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