The equivalent weight of a metal is 9 and vapour density of its chloride is 59.25. The atomic weight of metal is :

Options

A. 23.9
B. 27.3
C. 36.3
D. 48.3

Answer

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

### Step 1: Calculate the Molecular Weight of the Metal Chloride
The vapor density (VD) of the metal chloride is given as 59.25. We use the formula for molecular weight (MW) based on vapor density:
\[ \text{Molecular Weight} = \text{Vapor Density} \times 2 \]
Substituting the given value:
\[ \text{Molecular Weight} = 59.25 \times 2 = 118.5 \, \text{g/mol} \]

### Step 2: Identify the Atomic Weight of Chlorine
The atomic weight of chlorine (Cl) is known to be approximately:
\[ \text{Atomic Weight of Cl} = 35.5 \, \text{g/mol} \]

### Step 3: Use the Formula for Valency
The valency (n) can be calculated using the formula:
\[ \text{Valency} = \frac{\text{Molecular Weight of Metal Chloride}}{\text{Equivalent Weight of Metal} + \text{Atomic Weight of Chlorine}} \]
Substituting the known values:
\[ \text{Valency} = \frac{118.5}{9 + 35.5} = \frac{118.5}{44.5} \approx 2.66 \]

### Step 4: Relate Equivalent Weight, Atomic Weight, and Valency
The equivalent weight (E) of the metal is related to its atomic weight (A) and valency (n) by the formula:
\[ E = \frac{A}{n} \]
Rearranging this gives us:
\[ A = E \times n \]
Substituting the values we have:
\[ A = 9 \times 2.66 \approx 23.94 \, \text{g/mol} \]

### Final Answer
The atomic weight of the metal is approximately:
\[ \text{Atomic Weight of Metal} \approx 23.94 \, \text{g/mol} \]

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