In a metal chloride, the weight of metal and chlorine are in the ratio of `1 : 2`. The equivalent weight of the metal will be-

Options

A. 71
B. 35.5
C. 106.5
D. 17.75

Answer

To find the equivalent weight of the metal in a metal chloride where the weight of the metal and chlorine are in the ratio of 1:2, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ratio**:
Given that the weight of the metal (M) and chlorine (Cl) is in the ratio of 1:2, we can express this as:
\[ \text{Weight of Metal} : \text{Weight of Chlorine} = 1 : 2 \]
Let the weight of the metal be \( x \) and the weight of chlorine be \( 2x \).

2. **Determine the Equivalent Weight of Chlorine**:
The equivalent weight of an element is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Atomic Mass}}{\text{Valency}} \]
For chlorine (Cl), the atomic mass is approximately 35.5 and its valency in metal chlorides is -1 (since it typically forms Cl^- ions). Therefore, the equivalent weight of chlorine is:
\[ \text{Equivalent Weight of Cl} = \frac{35.5}{1} = 35.5 \]

3. **Set Up the Equation for Equivalent Weights**:
According to the problem, the number of gram equivalents of the metal will equal the number of gram equivalents of chlorine. This can be expressed as:
\[ \frac{\text{Weight of Metal}}{\text{Equivalent Weight of Metal}} = \frac{\text{Weight of Chlorine}}{\text{Equivalent Weight of Chlorine}} \]
Substituting the known values:
\[ \frac{x}{E_m} = \frac{2x}{35.5} \]
Here, \( E_m \) is the equivalent weight of the metal.

4. **Solve for the Equivalent Weight of the Metal**:
We can cancel \( x \) from both sides (assuming \( x \neq 0 \)):
\[ \frac{1}{E_m} = \frac{2}{35.5} \]
Rearranging gives:
\[ E_m = \frac{35.5}{2} = 17.75 \]

5. **Conclusion**:
The equivalent weight of the metal is:
\[ \boxed{17.75} \]