Vapour density of metal chloride is 77 Equivalent weight of Metal is 4, then its atomic weight will be :-

Options

A. 4
B. 16
C. 9
D. 12

Answer

To find the atomic weight of the metal from the given information, we can follow these steps:

### Step 1: Calculate the Molecular Weight
We know that the vapor density (VD) of the metal chloride is given as 77. The molecular weight (MW) can be calculated using the formula:
\[ \text{Molecular Weight} = 2 \times \text{Vapor Density} \]
Substituting the value:
\[ \text{Molecular Weight} = 2 \times 77 = 154 \]

### Step 2: Determine the Equivalent Weight of Metal Chloride
The equivalent weight of the metal chloride can be expressed as the sum of the equivalent weight of the metal and the equivalent weight of chlorine. The equivalent weight of chlorine is approximately 35.5. Given that the equivalent weight of the metal is 4, we can write:
\[ \text{Equivalent Weight of Metal Chloride} = \text{Equivalent Weight of Metal} + \text{Equivalent Weight of Chlorine} \]
Substituting the values:
\[ \text{Equivalent Weight of Metal Chloride} = 4 + 35.5 = 39.5 \]

### Step 3: Calculate the n-factor
The n-factor can be calculated using the formula:
\[ n = \frac{\text{Molecular Weight}}{\text{Equivalent Weight of Metal Chloride}} \]
Substituting the values we calculated:
\[ n = \frac{154}{39.5} \approx 3.89 \]
We can round this to approximately 4, as n-factors are typically whole numbers.

### Step 4: Calculate the Atomic Weight of the Metal
The atomic weight (AW) of the metal can be calculated using the formula:
\[ \text{Atomic Weight} = \text{Equivalent Weight} \times n \]
Substituting the values:
\[ \text{Atomic Weight} = 4 \times 4 = 16 \]

### Final Answer
Thus, the atomic weight of the metal is **16**.

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