When `6xx10^(22)` electrons are used in the electrolysis of a metalic salt, `1.6` gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So oxidation state of the metal in the salt is

Options

A. `+2`
B. `+3`
C. `+1`
D. `+4`

Explanation

To determine the oxidation state of the metal in the salt, we can follow these steps:

### Step 1: Calculate the number of moles of electrons
The number of electrons given is \(6 \times 10^{22}\). To find the number of moles of electrons, we use Avogadro's number (\(6.022 \times 10^{23}\)):

\[ \text{Number of moles of electrons} = \frac{6 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.1 \text{ moles} \]

### Step 2: Relate moles of metal deposited to moles of electrons
Let \(n\) be the oxidation state of the metal. The relationship between the moles of metal deposited and the moles of electrons can be expressed as:

\[ \text{Moles of metal} = \frac{\text{Moles of electrons}}{n} = \frac{0.1}{n} \]

### Step 3: Calculate the moles of metal from the mass deposited
We know that the mass of the metal deposited is \(1.6 \text{ g}\) and the atomic weight of the metal is \(57 \text{ g/mol}\). Thus, we can calculate the moles of metal deposited:

\[ \text{Moles of metal} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.6 \text{ g}}{57 \text{ g/mol}} \approx 0.0281 \text{ moles} \]

### Step 4: Set the equations equal to each other
Now we can set the equations for moles of metal equal to each other:

\[ \frac{0.1}{n} = 0.0281 \]

### Step 5: Solve for \(n\)
Rearranging the equation gives:

\[ n = \frac{0.1}{0.0281} \approx 3.56 \]

Since oxidation states are typically whole numbers, we can round this to \(n \approx 3\).

### Conclusion
The oxidation state of the metal in the salt is approximately \(+3\).

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