In what ratio should `._(17)CI^(37)` and `._(17)CI^(35)` be presents so as to obtain `._(17)CI^(35.5)`?

Options

A. `1:2`
B. `1:1`
C. `1:3`
D. `3:1`

Explanation

To solve the problem of finding the ratio in which the isotopes Cl-37 and Cl-35 should be present to obtain an average atomic mass of Cl-35.5, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
Let the mass of Cl-37 be represented as \( x \) and the mass of Cl-35 be represented as \( y \). We are looking for the ratio \( \frac{x}{y} \).

2. **Set Up the Equation for Average Atomic Mass**:
The average atomic mass can be calculated using the formula:
\[ \text{Average Atomic Mass} = \frac{(37 \cdot x) + (35 \cdot y)}{x + y} \]
We know that the average atomic mass we want is 35.5. Therefore, we can set up the equation:
\[ 35.5 = \frac{(37 \cdot x) + (35 \cdot y)}{x + y} \]

3. **Cross-Multiply to Eliminate the Denominator**:
By cross-multiplying, we get:
\[ 35.5(x + y) = 37x + 35y \]

4. **Expand and Rearrange the Equation**:
Expanding the left side gives:
\[ 35.5x + 35.5y = 37x + 35y \]
Rearranging the equation to isolate terms involving \( x \) and \( y \):
\[ 35.5y - 35y = 37x - 35.5x \]
This simplifies to:
\[ 0.5y = 1.5x \]

5. **Solve for the Ratio \( \frac{x}{y} \)**:
Dividing both sides by \( y \) and \( x \) gives:
\[ \frac{x}{y} = \frac{0.5}{1.5} = \frac{1}{3} \]

6. **Express the Ratio**:
Therefore, the ratio of Cl-37 to Cl-35 is:
\[ x:y = 1:3 \]

### Final Answer:
The ratio in which Cl-37 and Cl-35 should be present to obtain Cl-35.5 is **1:3**.