12.5 gm of fuming `H_(2)SO_(4)` (labelled as 112%) is mixed with 100 lit water. Molar concentration of `H^(+)` in resultant solution is
[Note: Assume that `H_(2)SO_(4)` dissociate completely and there is no change in volume on mixing

Options

A. `(2)/(700)`
B. `(2)/(350)`
C. `(3)/(350)`
D. `(3)/(700)`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the mass of `H₂SO₄` in the fuming solution
Given that the fuming `H₂SO₄` is labeled as 112%, this means that the solution contains 112 grams of `H₂SO₄` in 100 grams of solution.

To find the actual mass of `H₂SO₄` in 12.5 grams of the solution, we can set up a proportion:

\[ \text{Mass of } H₂SO₄ = \left( \frac{112 \text{ g}}{100 \text{ g solution}} \right) \times 12.5 \text{ g solution} \]

Calculating this gives:

\[ \text{Mass of } H₂SO₄ = \frac{112 \times 12.5}{100} = 14 \text{ g} \]

### Step 2: Calculate the number of moles of `H₂SO₄`
The molar mass of `H₂SO₄` is approximately 98 g/mol. We can calculate the number of moles using the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{14 \text{ g}}{98 \text{ g/mol}} \approx 0.1429 \text{ moles} \]

### Step 3: Determine the concentration of `H⁺` ions
When `H₂SO₄` dissociates completely, it produces 2 moles of `H⁺` ions for every mole of `H₂SO₄`. Therefore, the number of moles of `H⁺` produced is:

\[ \text{Moles of } H⁺ = 2 \times \text{moles of } H₂SO₄ = 2 \times 0.1429 \approx 0.2857 \text{ moles} \]

### Step 4: Calculate the molar concentration of `H⁺`
The total volume of the solution is 100 liters (assuming no change in volume upon mixing). The molar concentration (C) of `H⁺` ions can be calculated using the formula:

\[ C = \frac{\text{number of moles}}{\text{volume in liters}} = \frac{0.2857 \text{ moles}}{100 \text{ L}} = 0.002857 \text{ M} \]

### Step 5: Final Result
Thus, the molar concentration of `H⁺` in the resultant solution is approximately:

\[ C \approx 0.002857 \text{ M} \]