Vapour density of metal chloride is 77. Equivalent : weight of metal is 3, then its atomic weight will be

Options

A. 3
B. 6
C. 9
D. 12

Explanation

To find the atomic weight of the metal in the metal chloride given that the vapor density is 77 and the equivalent weight of the metal is 3, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between vapor density and molar mass**:
The vapor density (VD) is related to the molar mass (M) by the formula:
\[ \text{Vapor Density} = \frac{\text{Molar Mass}}{2} \]
Therefore, we can express the molar mass as:
\[ \text{Molar Mass} = \text{Vapor Density} \times 2 \]

2. **Calculate the molar mass of the metal chloride**:
Given that the vapor density of the metal chloride is 77, we can calculate the molar mass:
\[ \text{Molar Mass} = 77 \times 2 = 154 \, \text{g/mol} \]

3. **Use the equivalent weight to find the atomic weight of the metal**:
The equivalent weight (EW) is defined as:
\[ \text{Equivalent Weight} = \frac{\text{Atomic Weight}}{n} \]
where \( n \) is the number of moles of replaceable hydrogen or the valency of the metal in the compound.

In this case, we are given that the equivalent weight of the metal is 3. Assuming the metal chloride is of the form \( MCl_x \), the equivalent weight can also be expressed in terms of the molar mass of the metal chloride:
\[ \text{Equivalent Weight of Metal Chloride} = \frac{\text{Molar Mass of Metal Chloride}}{x} \]

4. **Relate the molar mass of the metal chloride to its components**:
The molar mass of the metal chloride can be expressed as:
\[ \text{Molar Mass of Metal Chloride} = \text{Atomic Weight of Metal} + x \times \text{Atomic Weight of Chlorine} \]
The atomic weight of chlorine (Cl) is approximately 35.5 g/mol.

5. **Set up the equation**:
Let \( A \) be the atomic weight of the metal. Then:
\[ 154 = A + x \times 35.5 \]

6. **Substituting the equivalent weight**:
From the equivalent weight formula:
\[ 3 = \frac{A}{x} \]
This implies:
\[ A = 3x \]

7. **Substituting \( A \) back into the molar mass equation**:
Substitute \( A \) in the molar mass equation:
\[ 154 = 3x + x \times 35.5 \]
Simplifying gives:
\[ 154 = x(3 + 35.5) \implies 154 = x \times 38.5 \]

8. **Solve for \( x \)**:
\[ x = \frac{154}{38.5} \approx 4 \]

9. **Find the atomic weight \( A \)**:
Now substituting \( x = 4 \) back into \( A = 3x \):
\[ A = 3 \times 4 = 12 \]

### Final Answer:
The atomic weight of the metal is **12 g/mol**.