The weight of solute present in 200mL of `0.1M H_(2)SO_(4)`:

Options

A. `2.45g`
B. `4.9g`
C. `1.96g`
D. `3.92g`

Explanation

To find the weight of solute present in 200 mL of 0.1 M H₂SO₄, we can follow these steps:

### Step 1: Understand the concept of molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:
\[ M = \frac{n}{V} \]
where:
- \( M \) = molarity (in moles per liter)
- \( n \) = number of moles of solute
- \( V \) = volume of solution in liters

### Step 2: Convert the volume from mL to L
Given that the volume of the solution is 200 mL, we need to convert this to liters:
\[ V = 200 \text{ mL} = 0.2 \text{ L} \]

### Step 3: Calculate the number of moles of H₂SO₄
Using the molarity formula, we can rearrange it to find the number of moles:
\[ n = M \times V \]
Substituting the values we have:
\[ n = 0.1 \text{ M} \times 0.2 \text{ L} = 0.02 \text{ moles} \]

### Step 4: Find the molar mass of H₂SO₄
The molar mass of H₂SO₄ (sulfuric acid) can be calculated as follows:
- Hydrogen (H) = 1.01 g/mol × 2 = 2.02 g/mol
- Sulfur (S) = 32.07 g/mol
- Oxygen (O) = 16.00 g/mol × 4 = 64.00 g/mol

Adding these together:
\[ \text{Molar mass of H₂SO₄} = 2.02 + 32.07 + 64.00 = 98.09 \text{ g/mol} \]

### Step 5: Calculate the weight of the solute
Now, we can use the number of moles and the molar mass to find the weight of the solute:
\[ \text{Weight of solute} = n \times \text{Molar mass} \]
Substituting the values:
\[ \text{Weight of solute} = 0.02 \text{ moles} \times 98.09 \text{ g/mol} = 1.9618 \text{ g} \]

### Step 6: Round the answer
Rounding this to two decimal places, we find:
\[ \text{Weight of solute} \approx 1.96 \text{ g} \]

### Final Answer
The weight of solute present in 200 mL of 0.1 M H₂SO₄ is approximately **1.96 grams**.

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