9.8 grams of `H_(2)SO_(4)` is present in two litres of a solution. The molarity of the solution is:

Explanation

To calculate the molarity of a solution containing 9.8 grams of \( H_2SO_4 \) in 2 liters of solution, follow these steps:

### Step 1: Write down the formula for molarity.
The formula for molarity (M) is given by:
\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

### Step 2: Calculate the number of moles of \( H_2SO_4 \).
To find the number of moles, use the formula:
\[ \text{Number of moles} = \frac{\text{Mass of solute (g)}}{\text{Molar mass of solute (g/mol)}} \]
The molar mass of \( H_2SO_4 \) (sulfuric acid) 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:
\[ \text{Molar mass of } H_2SO_4 = 2 + 32 + 64 = 98 \text{ g/mol} \]

Now, substitute the values into the moles formula:
\[ \text{Number of moles} = \frac{9.8 \text{ g}}{98 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the molarity.
Now that we have the number of moles, we can substitute it into the molarity formula:
\[ M = \frac{0.1 \text{ moles}}{2 \text{ L}} = 0.05 \text{ M} \]

### Conclusion:
The molarity of the solution is:
\[ \text{Molarity} = 0.05 \text{ M} \]

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