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 based on the given equivalent weight and vapor density of its chloride, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Equivalent weight of the metal (E) = 9 g/equiv
- Vapor density of the metal chloride (VD) = 59.25

2. **Calculate the Molecular Weight of the Metal Chloride:**
- The molecular weight (M) of the metal chloride can be calculated using the formula:
\[ M = 2 \times \text{Vapor Density} \]
- Substituting the given vapor density:
\[ M = 2 \times 59.25 = 118.5 \text{ g/mol} \]

3. **Calculate the Valency of the Metal:**
- The formula for calculating the valency (V) of the metal is:
\[ V = \frac{M}{E + M_{Cl}} \]
- Where \( M_{Cl} \) is the molecular weight of chlorine (approximately 35.5 g/mol).
- Substituting the values:
\[ V = \frac{118.5}{9 + 35.5} = \frac{118.5}{44.5} \]
- Performing the division:
\[ V \approx 2.66 \]

4. **Calculate the Atomic Weight of the Metal:**
- The atomic weight (A) of the metal can be calculated using the formula:
\[ A = E \times V \]
- Substituting the equivalent weight and valency:
\[ A = 9 \times 2.66 \]
- Performing the multiplication:
\[ A \approx 23.94 \text{ g/mol} \]

5. **Final Result:**
- The atomic weight of the metal is approximately 23.9 g/mol.

### Conclusion:
The atomic weight of the metal is **23.9 g/mol**.

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