Chlorine exists in two isotopic forms `Cl-37` and Cl-35 but its atomic mass is 35.5. this indicates the ratio of Cl-37 and Cl-35 is appromimately

Options

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

Explanation

To find the ratio of the isotopes Cl-37 and Cl-35 based on their relative abundances and the average atomic mass of chlorine, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables:**
- Let the relative abundance of Cl-37 be \( P_1 \) (in percentage).
- Let the relative abundance of Cl-35 be \( P_2 \) (in percentage).
- The mass of Cl-37 is \( m_1 = 37 \, \text{amu} \).
- The mass of Cl-35 is \( m_2 = 35 \, \text{amu} \).

2. **Establish Relationships:**
- Since the total abundance must equal 100%, we have:
\[ P_1 + P_2 = 100 \]
- From this, we can express \( P_2 \) in terms of \( P_1 \):
\[ P_2 = 100 - P_1 \]

3. **Use the Average Atomic Mass:**
- The average atomic mass of chlorine is given as 35.5 amu. The formula for average atomic mass is:
\[ \text{Average Atomic Mass} = \frac{(m_1 \cdot P_1) + (m_2 \cdot P_2)}{P_1 + P_2} \]
- Substituting the known values:
\[ 35.5 = \frac{(37 \cdot P_1) + (35 \cdot (100 - P_1))}{100} \]

4. **Simplify the Equation:**
- Multiply both sides by 100 to eliminate the denominator:
\[ 3550 = 37P_1 + 3500 - 35P_1 \]
- Combine like terms:
\[ 3550 = 2P_1 + 3500 \]
- Rearranging gives:
\[ 2P_1 = 3550 - 3500 \]
\[ 2P_1 = 50 \]
\[ P_1 = 25 \]

5. **Calculate \( P_2 \):**
- Now substitute \( P_1 \) back to find \( P_2 \):
\[ P_2 = 100 - P_1 = 100 - 25 = 75 \]

6. **Find the Ratio:**
- The ratio of Cl-37 to Cl-35 is:
\[ \text{Ratio} = \frac{P_1}{P_2} = \frac{25}{75} = \frac{1}{3} \]

### Conclusion:
The approximate ratio of Cl-37 to Cl-35 is **1:3**.