The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:

Options

A. 70g
B. 35g
C. 30g
D. 95gm

Answer

To find the mass of 70% H₂SO₄ required for the neutralization of one mole of NaOH, we can follow these steps:

### Step 1: Determine the reaction between NaOH and H₂SO₄
The balanced chemical equation for the neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄) is:

\[ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \]

From the equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, to neutralize 1 mole of NaOH, we need:

\[ \text{H}_2\text{SO}_4 \text{ required} = \frac{1 \text{ mole NaOH}}{2} = 0.5 \text{ moles of H}_2\text{SO}_4 \]

### Step 2: Calculate the mass of pure H₂SO₄ needed
Next, we need to find the molar mass of H₂SO₄. The molar mass can be calculated as follows:

- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Sulfur (S): 32 g/mol × 1 = 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together gives:

\[ \text{Molar mass of H}_2\text{SO}_4 = 2 + 32 + 64 = 98 \text{ g/mol} \]

Now, we can calculate the mass of pure H₂SO₄ required for 0.5 moles:

\[ \text{Mass of H}_2\text{SO}_4 = 0.5 \text{ moles} \times 98 \text{ g/mol} = 49 \text{ g} \]

### Step 3: Calculate the mass of 70% H₂SO₄ solution needed
Since the H₂SO₄ solution is 70% by mass, we need to find out how much of this solution contains 49 g of pure H₂SO₄.

Let \( x \) be the mass of the 70% H₂SO₄ solution required:

\[ 0.70x = 49 \text{ g} \]

To find \( x \):

\[ x = \frac{49 \text{ g}}{0.70} = 70 \text{ g} \]

### Conclusion
The mass of 70% H₂SO₄ required for the neutralization of one mole of NaOH is **70 g**.

---