When 110 g of manganese (At mass =55) dissolves in dilute `HNO_(3)` at `27^(@)C` under atmosphere pressure, the work done in the process is

Options

A. `2494.2J`
B. `-2494.2J`
C. `-4988.4J`
D. `4988.4J`

Explanation

To solve the problem of calculating the work done when 110 g of manganese dissolves in dilute HNO₃, we can follow these steps:

### Step 1: Write the Chemical Reaction
The reaction of manganese (Mn) with dilute nitric acid (HNO₃) can be represented as:
\[ \text{Mn} + 2 \text{HNO}_3 \rightarrow \text{Mn(NO}_3\text{)}_2 + \text{H}_2 \]
This indicates that manganese reacts with nitric acid to form manganese nitrate and hydrogen gas.

### Step 2: Calculate the Number of Moles of Manganese
To find the number of moles of manganese, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of manganese = 110 g
- Molar mass of manganese = 55 g/mol

Calculating the number of moles:
\[ \text{Number of moles of Mn} = \frac{110 \text{ g}}{55 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Determine the Change in Moles of Gas (Δn)
From the balanced equation, we see that:
- 2 moles of Mn produce 2 moles of H₂ gas.
- There are no gaseous reactants.

Thus, the change in moles of gas (Δn) is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 0 = 2 \]

### Step 4: Calculate the Work Done (w)
The work done during the expansion can be calculated using the formula:
\[ w = -P \Delta V \]
Using the ideal gas law, we can express this in terms of Δn:
\[ w = -\Delta nRT \]
Where:
- R = 8.314 J/(mol·K) (universal gas constant)
- T = 27°C = 300 K (temperature in Kelvin)

Substituting the values:
\[ w = -2 \times 8.314 \text{ J/(mol·K)} \times 300 \text{ K} \]
\[ w = -4988.4 \text{ J} \]

### Step 5: Conclusion
The work done in the process is:
\[ \text{Work done} = -4988.4 \text{ J} \]