If `100 mL "of" 1 M H_(2) SO_(4)` solution is mixed with `100 mL` of 98% `(W//W)` of `H_(2)SO_(4)` solution `(d = 0.1 g mL^(-1))`, then

Options

A. Concentration of solution becomes half.
B. Volume of solution beomces `200 mL`.
C. Mass of `H_(2)SO_(4)` is the solution is `98 g`
D. mass of `H_(2)SO_(4)` in the solution is `19.6 g`.

Explanation

To solve the problem step by step, we will analyze the two solutions given and determine the molarity of the resulting mixture.

### Step 1: Calculate the moles of H₂SO₄ in the first solution

The first solution is 100 mL of 1 M H₂SO₄.

\[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Moles of H₂SO₄} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} \]

### Step 2: Calculate the moles of H₂SO₄ in the second solution

The second solution is 100 mL of 98% (W/W) H₂SO₄ with a density of 1.1 g/mL.

1. Calculate the mass of the solution:
\[ \text{Mass of solution} = \text{Volume} \times \text{Density} = 100 \, \text{mL} \times 1.1 \, \text{g/mL} = 110 \, \text{g} \]

2. Calculate the mass of H₂SO₄ in the solution:
\[ \text{Mass of H₂SO₄} = \text{Weight \%} \times \text{Mass of solution} = 0.98 \times 110 \, \text{g} = 107.8 \, \text{g} \]

3. Convert mass of H₂SO₄ to moles:
\[ \text{Moles of H₂SO₄} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{107.8 \, \text{g}}{98 \, \text{g/mol}} \approx 1.1 \, \text{mol} \]

### Step 3: Calculate the total moles of H₂SO₄ in the mixture

\[ \text{Total moles of H₂SO₄} = \text{Moles from first solution} + \text{Moles from second solution} \]
\[ \text{Total moles of H₂SO₄} = 0.1 \, \text{mol} + 1.1 \, \text{mol} = 1.2 \, \text{mol} \]

### Step 4: Calculate the total volume of the mixture

The total volume of the mixture is:
\[ \text{Total volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

### Step 5: Calculate the molarity of the resulting mixture

\[ \text{Molarity} = \frac{\text{Total moles}}{\text{Total volume (in L)}} \]
\[ \text{Molarity} = \frac{1.2 \, \text{mol}}{0.2 \, \text{L}} = 6 \, \text{M} \]

### Conclusion

The resulting mixture has a molarity of 6 M H₂SO₄.