`H_(2)SO_(4)` solution `(80% "by weight and specific gravity" 1.75 g//ml)` is used to prepare `2 litre` of `0.25 MH_(2)SO_(4)` (aq). The volume of `H_(2)SO_(4)` solution (original) which must be used is :

Options

A. `107.18ml`
B. `43.75ml`
C. `35ml`
D. None of these

Explanation

To solve the problem, we need to determine the volume of the `H₂SO₄` solution (80% by weight and specific gravity 1.75 g/mL) required to prepare 2 liters of a 0.25 M `H₂SO₄` solution.

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of `H₂SO₄` required for the final solution.**

The formula for calculating the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]

Given:
- Molarity (M) = 0.25 M
- Volume (V) = 2 L

Calculating the number of moles:
\[ \text{Number of moles} = 0.25 \, \text{mol/L} \times 2 \, \text{L} = 0.5 \, \text{mol} \]

**Step 2: Calculate the mass of `H₂SO₄` required.**

The molar mass of `H₂SO₄` is approximately 98 g/mol. Therefore, the mass required is:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass} = 0.5 \, \text{mol} \times 98 \, \text{g/mol} = 49 \, \text{g} \]

**Step 3: Calculate the volume of the `H₂SO₄` solution needed to obtain this mass.**

The solution is 80% by weight, meaning that in 100 g of solution, there are 80 g of `H₂SO₄`.

Let \( x \) be the mass of the solution needed. Then:
\[ 0.8x = 49 \, \text{g} \]

Solving for \( x \):
\[ x = \frac{49}{0.8} = 61.25 \, \text{g} \]

**Step 4: Convert the mass of the solution to volume using specific gravity.**

Specific gravity (SG) is given as 1.75 g/mL, which means that 1 mL of the solution weighs 1.75 g.

To find the volume \( V \):
\[ V = \frac{\text{mass}}{\text{specific gravity}} = \frac{61.25 \, \text{g}}{1.75 \, \text{g/mL}} \approx 35 \, \text{mL} \]

### Final Answer:
The volume of the `H₂SO₄` solution that must be used is approximately **35 mL**.

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