The molarity of 0.5 N `H_(2)SO_(4)` solution is…………….

Answer

To find the molarity of a 0.5 N `H₂SO₄` (sulfuric acid) solution, we can use the relationship between normality and molarity. Here’s a step-by-step solution:

### Step 1: Understand the Relationship
The relationship between normality (N) and molarity (M) is given by the formula:
\[ \text{Normality} = n \times \text{Molarity} \]
where \( n \) is the number of equivalents.

### Step 2: Determine the n Factor
For sulfuric acid (`H₂SO₄`), it can donate 2 protons (H⁺ ions) in solution. Therefore, the n factor for sulfuric acid is 2.

### Step 3: Substitute Known Values
We know:
- Normality (N) = 0.5 N
- n factor = 2

Using the formula:
\[ 0.5 = 2 \times \text{Molarity} \]

### Step 4: Solve for Molarity
Rearranging the equation to solve for molarity:
\[ \text{Molarity} = \frac{0.5}{2} \]

Calculating this gives:
\[ \text{Molarity} = 0.25 \, \text{M} \]

### Final Answer
The molarity of the 0.5 N `H₂SO₄` solution is **0.25 M**.

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