When an acidified solution of `Na_2MoO_n` (atomic mass of Moi=36) is electrolyzed, `O_2` gas is liberated corresponding to a volume of 0.112 L at STP and mass of MO deposited is 0.32 g. Then the formula of the salt and oxidation state of Mo is

Options

A. `Na_2MoO`,0
B. `Na_2MoO_4`,+6
C. `Na_2MoO_2`,+2
D. `Na_2MoO_3`,+4

Answer

To solve the problem, we need to determine the formula of the salt `Na2MoOn` and the oxidation state of molybdenum (Mo) based on the given information.

### Step 1: Calculate the mass of oxygen liberated
We know that the volume of oxygen gas liberated at STP is 0.112 L. To find the mass of oxygen, we can use the molar volume of a gas at STP, which is 22.4 L/mol.

\[ \text{Mass of } O_2 = \left( \frac{0.112 \, \text{L}}{22.4 \, \text{L/mol}} \right) \times 32 \, \text{g/mol} \]

Calculating this gives:

\[ \text{Mass of } O_2 = \left( \frac{0.112}{22.4} \right) \times 32 = 0.16 \, \text{g} \]

### Step 2: Use Faraday's second law of electrolysis
According to Faraday's second law, the mass of a substance deposited during electrolysis is proportional to the equivalent weight of the substance and the amount of electricity passed.

Let the equivalent weight of molybdenum (Mo) be \( \frac{M}{n} \), where \( M \) is the molar mass of Mo (96 g/mol) and \( n \) is the number of electrons transferred.

The equivalent weight of oxygen is 8 g (since O2 has a molar mass of 32 g and it takes 4 electrons to reduce O2 to 2O2-).

### Step 3: Set up the ratio of masses
Using the mass of oxygen and molybdenum deposited, we can set up the ratio:

\[ \frac{\text{Mass of } O_2}{\text{Mass of Mo}} = \frac{\text{Equivalent weight of } O_2}{\text{Equivalent weight of Mo}} \]

Substituting the known values:

\[ \frac{0.16 \, \text{g}}{0.32 \, \text{g}} = \frac{8}{\frac{96}{n}} \]

### Step 4: Solve for n
Cross-multiplying gives:

\[ 0.16 \times \frac{96}{n} = 0.32 \times 8 \]

Calculating the right side:

\[ 0.32 \times 8 = 2.56 \]

Thus, we have:

\[ 0.16 \times 96 = 2.56n \]

Calculating the left side:

\[ 15.36 = 2.56n \]

Now, solving for \( n \):

\[ n = \frac{15.36}{2.56} = 6 \]

### Step 5: Determine the oxidation state of Mo
The oxidation state of Mo in the salt can be determined from the value of \( n \):

\[ \text{Oxidation state of Mo} = +6 \]

### Step 6: Write the formula of the salt
The formula of the salt can be deduced as follows:

- Sodium (Na) has an oxidation state of +1.
- Molybdenum (Mo) has an oxidation state of +6.
- Oxygen (O) has an oxidation state of -2.

To balance the charges in the salt formula \( Na_2MoO_n \):

\[ 2(+1) + (+6) + n(-2) = 0 \]

This simplifies to:

\[ 2 + 6 - 2n = 0 \implies 8 = 2n \implies n = 4 \]

Thus, the formula of the salt is:

\[ Na_2MoO_4 \]

### Final Answer
The formula of the salt is \( Na_2MoO_4 \) and the oxidation state of Mo is +6.