The equiv. wt. of an element is 9. If it forms volatile chloride of vapour density 58.5. What is the approximate at wt. of the element ?

Options

A. 9
B. 18
C. 27
D. 54

Explanation

To solve the problem step by step, follow the instructions below:

### Step 1: Understand the Given Information
We are given:
- Equivalent weight of the element (E) = 9
- Vapor density (VD) of the volatile chloride = 58.5

### Step 2: Use the Vapor Density Formula
The formula for vapor density (VD) is given by:
\[ VD = \frac{Molecular \ Weight}{2} \]
From this, we can rearrange the formula to find the molecular weight (MW):
\[ Molecular \ Weight = Vapor \ Density \times 2 \]
Substituting the given vapor density:
\[ Molecular \ Weight = 58.5 \times 2 = 117 \]

### Step 3: Set Up the Molecular Weight Equation
Let the atomic weight of the element be \( M \). The volatile chloride formed will be \( MCl_x \), where \( x \) is the valency of the element. The molecular weight of the chloride can be expressed as:
\[ Molecular \ Weight = M + x \times 35.5 \]
Here, 35.5 is the atomic weight of chlorine.

### Step 4: Equate the Molecular Weights
From the previous steps, we have:
\[ M + x \times 35.5 = 117 \]

### Step 5: Relate Equivalent Weight to Molecular Weight
The equivalent weight (E) of the element is related to its molecular weight and valency (x) by the formula:
\[ E = \frac{M}{x} \]
We know that \( E = 9 \), so:
\[ 9 = \frac{M}{x} \implies M = 9x \]

### Step 6: Substitute and Solve for x
Now substitute \( M = 9x \) into the molecular weight equation:
\[ 9x + x \times 35.5 = 117 \]
Combine like terms:
\[ x(9 + 35.5) = 117 \]
\[ x(44.5) = 117 \]
Now solve for \( x \):
\[ x = \frac{117}{44.5} \approx 2.63 \]
Rounding \( x \) gives \( x \approx 3 \).

### Step 7: Calculate the Atomic Weight of the Element
Now that we have \( x \), we can find \( M \):
\[ M = 9x = 9 \times 3 = 27 \]

### Conclusion
The approximate atomic weight of the element is **27**.