Molarity of 0.1 N oxalic acids

Options

A. 0.05 M
B. 0.1 M
C. 0.2 M
D. 0.3 M

Explanation

To find the molarity of 0.1 N oxalic acid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Normality and Molarity**:
The relationship between normality (N) and molarity (M) is given by the formula:
\[ N = n \times M \]
where \( n \) is the valency factor (the number of equivalent ions produced by one molecule of the solute).

2. **Identify the Valency Factor for Oxalic Acid**:
Oxalic acid (H₂C₂O₄) has two dissociable hydrogen ions (H⁺) in its structure. Therefore, the valency factor \( n \) for oxalic acid is:
\[ n = 2 \]

3. **Rearrange the Formula to Find Molarity**:
We can rearrange the formula to solve for molarity:
\[ M = \frac{N}{n} \]

4. **Substitute the Values**:
Now substitute the values of normality (0.1 N) and the valency factor (2) into the equation:
\[ M = \frac{0.1}{2} \]

5. **Calculate the Molarity**:
Performing the calculation gives:
\[ M = 0.05 \, \text{mol/L} \]

### Final Answer:
The molarity of 0.1 N oxalic acid is **0.05 mol/L**.

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