The amount of water produced (in g) in the oxidation of 1 mol of rhombic sulphur by conc. `HNO_3` to a compound with the highest oxidation state of sulphur is………….. (Given data: Molar mass of water = 18 g mol`""^(-1))`

Explanation

To solve the problem of determining the amount of water produced when 1 mole of rhombic sulfur (S8) is oxidized by concentrated nitric acid (HNO3) to a compound with the highest oxidation state of sulfur, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is rhombic sulfur (S8).
- The oxidizing agent is concentrated nitric acid (HNO3).
- The products of this reaction include sulfuric acid (H2SO4), nitrogen dioxide (NO2), and water (H2O).

2. **Write the Balanced Chemical Equation**:
- The balanced equation for the oxidation of rhombic sulfur by concentrated HNO3 can be written as:
\[ S_8 + 48 HNO_3 \rightarrow 8 H_2SO_4 + 48 NO_2 + 16 H_2O \]

3. **Determine the Moles of Water Produced**:
- From the balanced equation, we see that 1 mole of S8 produces 16 moles of water (H2O).

4. **Calculate the Mass of Water Produced**:
- To find the mass of water produced, we use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
- The molar mass of water (H2O) is calculated as follows:
\[ \text{Molar mass of H2O} = 2 \times 1 + 16 = 18 \, \text{g/mol} \]
- Therefore, the mass of water produced from 16 moles is:
\[ \text{Mass of water} = 16 \, \text{moles} \times 18 \, \text{g/mol} = 288 \, \text{g} \]

5. **Final Answer**:
- The amount of water produced in grams when 1 mole of rhombic sulfur is oxidized by concentrated HNO3 is **288 g**.