A current of 5.0 A flows for 4.0 h through an electrolytic cell containing a molten slat of metal M. This result in deposition of 0.25 mol of the metal M at the cathode. The oxidation state of M in the molten salt is `+x`. The value of 'x' is (1 Faraday = 96000 C `mol^(-1)`)

Answer

To solve the problem, we need to determine the oxidation state \( x \) of the metal \( M \) in the molten salt \( M^x \). We know that a current of 5.0 A flows for 4.0 hours, resulting in the deposition of 0.25 moles of metal \( M \) at the cathode.

### Step-by-Step Solution:

1. **Convert Time to Seconds**:
\[ \text{Time (t)} = 4.0 \text{ hours} = 4.0 \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 14400 \text{ seconds} \]

2. **Calculate Total Charge (Q)**:
Using the formula \( Q = I \times t \):
\[ Q = 5.0 \text{ A} \times 14400 \text{ s} = 72000 \text{ C} \]

3. **Relate Charge to Moles of Electrons**:
According to Faraday's law, the number of moles of electrons (\( n \)) can be calculated using:
\[ n = \frac{Q}{F} \]
where \( F = 96000 \text{ C/mol} \) (Faraday's constant).
\[ n = \frac{72000 \text{ C}}{96000 \text{ C/mol}} = 0.75 \text{ mol of electrons} \]

4. **Determine the Number of Electrons Transferred per Mole of Metal**:
Since 0.25 moles of metal \( M \) are deposited, and \( n \) moles of electrons are required to deposit 1 mole of metal, we can set up the equation:
\[ 0.25 \text{ mol of } M \text{ requires } 0.75 \text{ mol of electrons} \]
Therefore, for 1 mole of \( M \):
\[ n_e = \frac{0.75 \text{ mol}}{0.25 \text{ mol}} = 3 \]
This means that 3 moles of electrons are required to deposit 1 mole of metal \( M \).

5. **Determine the Oxidation State \( x \)**:
Since the metal ion \( M^x \) gains electrons to become neutral metal \( M \), we have:
\[ x = n_e = 3 \]

Thus, the oxidation state \( x \) of the metal \( M \) in the molten salt is **+3**.

### Final Answer:
The value of \( x \) is **3**.