Weight of solute present in 500 ml 0.2 `N-H_(2)SO_(4)` solution is

Options

A. 14.2 g
B. 4.9 g
C. 3.55 g
D. 1.42 g

Explanation

To find the weight of solute present in a 500 ml 0.2 N H₂SO₄ solution, we can follow these steps:

### Step 1: Understand the relationship between normality, volume, and equivalents
Normality (N) is defined as the number of equivalents of solute per liter of solution. The formula for calculating the number of equivalents is:
\[ \text{Number of Equivalents} = \text{Normality} \times \text{Volume (in liters)} \]

### Step 2: Convert the volume from ml to liters
Given that the volume of the solution is 500 ml, we convert it to liters:
\[ \text{Volume in liters} = \frac{500 \text{ ml}}{1000} = 0.5 \text{ L} \]

### Step 3: Calculate the number of equivalents
Using the normality and the volume in liters:
\[ \text{Number of Equivalents} = 0.2 \, \text{N} \times 0.5 \, \text{L} = 0.1 \, \text{equivalents} \]

### Step 4: Determine the n-factor for H₂SO₄
The n-factor (valency factor) for sulfuric acid (H₂SO₄) is 2 because it can donate two protons (H⁺ ions) in solution.

### Step 5: Calculate the molar mass of H₂SO₄
The molar mass of H₂SO₄ can be calculated as follows:
- Hydrogen (H): 2 atoms × 1 g/mol = 2 g/mol
- Sulfur (S): 1 atom × 32 g/mol = 32 g/mol
- Oxygen (O): 4 atoms × 16 g/mol = 64 g/mol

Total molar mass:
\[ \text{Molar mass of H₂SO₄} = 2 + 32 + 64 = 98 \, \text{g/mol} \]

### Step 6: Calculate the weight of the solute
Using the formula:
\[ \text{Weight of solute} = \text{Number of Equivalents} \times \text{n-factor} \times \text{Molar mass} \]
Substituting the known values:
\[ \text{Weight of solute} = 0.1 \, \text{equivalents} \times 2 \times 98 \, \text{g/mol} \]
\[ \text{Weight of solute} = 0.1 \times 2 \times 98 = 19.6 \, \text{grams} \]

### Conclusion
The weight of solute present in 500 ml of 0.2 N H₂SO₄ solution is **19.6 grams**.

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