PASSAGE-IV : 10 g of NaOH required certain amount of `H_(2)SO_(4)` for complete neutralisation.
Calculate the number of mole of `H_(2)SO_(4)` in it.

Options

A. 1 mole
B. 0.1 mole
C. 0.5 mole
D. 0.125 mole

Explanation

To solve the problem of how many moles of \( H_2SO_4 \) are required to completely neutralize 10 g of \( NaOH \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄) is given by the equation:
\[ 2 \, \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} \]

### Step 2: Determine the molar mass of NaOH
To find the number of moles of \( NaOH \), we first need to calculate its molar mass:
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol

Thus, the molar mass of \( NaOH \) is:
\[ \text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of NaOH
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

We can calculate the number of moles of \( NaOH \):
\[ \text{Number of moles of NaOH} = \frac{10 \, \text{g}}{40 \, \text{g/mol}} = 0.25 \, \text{moles} \]

### Step 4: Use the stoichiometry of the reaction
From the balanced equation, we see that 2 moles of \( NaOH \) react with 1 mole of \( H_2SO_4 \). Therefore, we can set up a proportion to find the moles of \( H_2SO_4 \) required:
\[ 2 \, \text{moles NaOH} \rightarrow 1 \, \text{mole H}_2\text{SO}_4 \]

Thus, for 0.25 moles of \( NaOH \):
\[ \text{Moles of } H_2SO_4 = \frac{1}{2} \times 0.25 = 0.125 \, \text{moles} \]

### Conclusion
The number of moles of \( H_2SO_4 \) required for the complete neutralization of 10 g of \( NaOH \) is:
\[ \text{0.125 moles} \]