The vapour density of chloride of an element is 39.5. The `Ew` of the elements is 3.82. The atomic weight of the elements is

Options

A. 15.28
B. 7.64
C. 3.82
D. 11.46

Explanation

To find the atomic weight of the element based on the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Vapor Density**:
- The vapor density (VD) of the chloride of the element is given as 39.5.
- The relationship between vapor density and molar mass (molecular weight) is given by the formula:
\[ \text{Molar Mass} = 2 \times \text{Vapor Density} \]
- Therefore, we can calculate the molar mass of the chloride:
\[ \text{Molar Mass} = 2 \times 39.5 = 79 \text{ g/mol} \]

2. **Setting up the Molecular Weight Equation**:
- Let the element be represented as \(M\) and the chloride as \(MCl_x\), where \(x\) is the number of chlorine atoms.
- The molar mass of the chloride can be expressed as:
\[ M + x \times 35.5 = 79 \]
- Here, 35.5 g/mol is the atomic weight of chlorine.

3. **Using Equivalent Weight**:
- The equivalent weight of the element is given as 3.82.
- The relationship between atomic weight (A), equivalent weight (Ew), and valency (x) is:
\[ A = \text{Ew} \times x \]
- Thus, we can express the atomic weight as:
\[ A = 3.82 \times x \]

4. **Substituting and Solving for x**:
- From the molar mass equation:
\[ M = 79 - x \times 35.5 \]
- Substituting \(M\) in the atomic weight equation:
\[ 79 - x \times 35.5 = 3.82 \times x \]
- Rearranging gives:
\[ 79 = x \times 35.5 + 3.82 \times x \]
\[ 79 = x(35.5 + 3.82) \]
\[ 79 = x(39.32) \]
- Solving for \(x\):
\[ x = \frac{79}{39.32} \approx 2 \]

5. **Finding the Atomic Weight**:
- Now that we have \(x\), we can find the atomic weight:
\[ A = 3.82 \times 2 = 7.64 \]

### Final Answer:
The atomic weight of the element is approximately **7.64 g/mol**.