`M^(2+) +2e rarr M. 0.275 g` of metal `M` is deposited at the cathode due to passage of `1A` of current for `965s.` Hence atomic weight of the metal `M` is:-

Options

A. `27.5`
B. `55.0`
C. `110.0`
D. `13.75`

Explanation

To find the atomic weight of the metal \( M \) deposited at the cathode, we can use Faraday's laws of electrolysis. Here's a step-by-step solution:

### Step 1: Identify the Given Data
- Mass of metal deposited (\( W \)) = 0.275 g
- Current (\( I \)) = 1 A
- Time (\( t \)) = 965 s
- Faraday's constant (\( F \)) = 96500 C/mol
- Valency factor (\( n \)) = 2 (since \( M^{2+} + 2e^- \rightarrow M \))

### Step 2: Calculate the Total Charge (Q)
Using the formula for charge:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 1 \, \text{A} \times 965 \, \text{s} = 965 \, \text{C} \]

### Step 3: Use Faraday's Law
According to Faraday's law:
\[ \frac{W}{E} = \frac{Q}{F} \]
Where:
- \( W \) = mass of the substance deposited
- \( E \) = equivalent weight of the substance
- \( Q \) = total charge
- \( F \) = Faraday's constant

Rearranging gives:
\[ E = \frac{W \times F}{Q} \]

### Step 4: Substitute the Values to Find Equivalent Weight
Substituting the values we have:
\[ E = \frac{0.275 \, \text{g} \times 96500 \, \text{C/mol}}{965 \, \text{C}} \]

Calculating this:
\[ E = \frac{0.275 \times 96500}{965} = 0.275 \times 100 = 27.5 \, \text{g/mol} \]

### Step 5: Relate Equivalent Weight to Atomic Weight
The equivalent weight \( E \) is related to the atomic weight \( A \) by the formula:
\[ E = \frac{A}{n} \]
Where \( n \) is the valency factor (which is 2 in this case).

Rearranging gives:
\[ A = E \times n \]
Substituting the values:
\[ A = 27.5 \, \text{g/mol} \times 2 = 55 \, \text{g/mol} \]

### Final Answer
The atomic weight of the metal \( M \) is \( 55 \, \text{g/mol} \).

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