Sodium nitrate on reduction with Zn in presence of NaOH solution produces `NH_(3)`. Mass of sodium nitrate absorbing 1 mole of electron will be .

Options

A. 7.75
B. 10.625
C. 8
D. 9.875

Explanation

To solve the problem, we need to determine the mass of sodium nitrate (NaNO3) that absorbs 1 mole of electrons during its reduction to ammonia (NH3) when reacted with zinc (Zn) in the presence of sodium hydroxide (NaOH).

### Step-by-Step Solution:

1. **Identify the Reaction**:
Sodium nitrate (NaNO3) is reduced to ammonia (NH3) in the presence of zinc and sodium hydroxide. The balanced half-reaction for the reduction of nitrate to ammonia can be represented as:
\[ \text{NO}_3^- + 8 \text{H}^+ + 8 \text{e}^- \rightarrow \text{NH}_3 + 3 \text{H}_2\text{O} \]
This indicates that 1 mole of NaNO3 (which contains 1 mole of NO3^-) absorbs 8 moles of electrons to produce 1 mole of NH3.

2. **Determine the Molar Mass of Sodium Nitrate**:
The molar mass of sodium nitrate (NaNO3) can be calculated as follows:
- Sodium (Na): 23 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of NaNO}_3 = 23 + 14 + 48 = 85 \text{ g/mol} \]

3. **Calculate the Mass of Sodium Nitrate that Absorbs 1 Mole of Electrons**:
Since 1 mole of NaNO3 absorbs 8 moles of electrons, the mass of NaNO3 that absorbs 1 mole of electrons can be calculated using the proportion:
\[ \text{Mass of NaNO}_3 \text{ for 1 mole of e}^- = \frac{85 \text{ g/mol}}{8} = 10.625 \text{ g} \]

4. **Final Answer**:
Therefore, the mass of sodium nitrate that absorbs 1 mole of electrons is **10.625 grams**.