`13g` of metal `M` is deposited at cathode by passing `0.4F` of electricity . The cathodic reaction is `:`
`M^(n+)+n e^(-) rarr M`
The formula of metal chloride `(Aw=65)` is `……………………..` .

Explanation

To solve the problem, we need to determine the formula of the metal chloride based on the given information. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- Mass of metal \( M \) deposited = 13 g
- Amount of electricity passed = 0.4 F (Faraday)
- Atomic weight of metal \( M \) = 65 g/mol
- Cathodic reaction: \( M^{n+} + n e^- \rightarrow M \)

### Step 2: Relate the amount of electricity to the mass of metal deposited
From Faraday's laws of electrolysis, we know that:
\[ \text{Mass of metal deposited} = \frac{(Atomic weight) \times (Charge)}{(n \times F)} \]

Where:
- \( F \) = Faraday constant (approximately 96500 C/mol)
- \( n \) = number of moles of electrons transferred

### Step 3: Set up the equation
Using the information provided:
\[ 13 \, \text{g} = \frac{65 \, \text{g/mol} \times 0.4 \, \text{F}}{n} \]

### Step 4: Solve for \( n \)
Rearranging the equation to solve for \( n \):
\[ n = \frac{65 \, \text{g/mol} \times 0.4 \, \text{F}}{13 \, \text{g}} \]

Calculating this:
\[ n = \frac{26}{13} = 2 \]

### Step 5: Determine the oxidation state of metal \( M \)
Since \( n = 2 \), the oxidation state of metal \( M \) is \( +2 \) (i.e., \( M^{2+} \)).

### Step 6: Determine the formula of the metal chloride
Chlorine has an oxidation state of \( -1 \). Therefore, to balance the charges:
- For \( M^{2+} \), we need 2 chloride ions \( Cl^- \) to balance the charge.

Thus, the formula for the metal chloride will be:
\[ \text{Metal Chloride} = MCl_2 \]

### Final Answer
The formula of the metal chloride is \( MCl_2 \).

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