Equivalent weight of a bivalent metal is `37.2`. The molecular weight of its chloride is

Options

A. (a)`412.2`
B. (b)`216`
C. (c )`145.4`
D. (d)`108.2`

Explanation

To find the molecular weight of the chloride of a bivalent metal with an equivalent weight of 37.2, we can follow these steps:

### Step 1: Understand the concept of equivalent weight
The equivalent weight of an element is defined as the atomic weight divided by its valency. For a bivalent metal, the valency is 2.

### Step 2: Calculate the atomic weight of the metal
Using the formula for equivalent weight:
\[ \text{Equivalent Weight} = \frac{\text{Atomic Weight}}{\text{Valency}} \]
We can rearrange this to find the atomic weight:
\[ \text{Atomic Weight} = \text{Equivalent Weight} \times \text{Valency} \]
Substituting the known values:
\[ \text{Atomic Weight} = 37.2 \, \text{g/mol} \times 2 = 74.4 \, \text{g/mol} \]

### Step 3: Determine the formula of the metal chloride
Since the metal is bivalent (valency = 2) and chlorine has a valency of 1, the formula for the metal chloride will be:
\[ \text{Metal Chloride} = \text{MCl}_2 \]

### Step 4: Calculate the molecular weight of the metal chloride
The molecular weight of the metal chloride can be calculated as follows:
\[ \text{Molecular Weight} = \text{Atomic Weight of Metal} + 2 \times \text{Atomic Weight of Chlorine} \]
The atomic weight of chlorine is approximately 35.5 g/mol. Therefore:
\[ \text{Molecular Weight} = 74.4 \, \text{g/mol} + 2 \times 35.5 \, \text{g/mol} \]
Calculating this gives:
\[ \text{Molecular Weight} = 74.4 + 71 = 145.4 \, \text{g/mol} \]

### Conclusion
The molecular weight of the metal chloride is **145.4 g/mol**.

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