How many grams of NaOH are equivalent to 100 ml of 0.1 N oxalic acid

Options

A. 0.2
B. 2
C. 0.4
D. 4

Explanation

To solve the problem of how many grams of NaOH are equivalent to 100 ml of 0.1 N oxalic acid, we can follow these steps:

### Step 1: Understand the relationship between normality, volume, and equivalent weight.
The formula for normality (N) is given by:
\[ N = \frac{W}{E \times V} \]
Where:
- \( N \) = Normality
- \( W \) = Weight of the solute (in grams)
- \( E \) = Equivalent weight of the solute (in grams/equiv)
- \( V \) = Volume of the solution (in liters)

### Step 2: Convert the volume from ml to liters.
Given that the volume of oxalic acid is 100 ml, we convert this to liters:
\[ V = \frac{100 \, \text{ml}}{1000} = 0.1 \, \text{L} \]

### Step 3: Calculate the equivalent weight of NaOH.
The equivalent weight of NaOH can be calculated as follows:
- The molar mass of Na (Sodium) = 23 g/mol
- The molar mass of O (Oxygen) = 16 g/mol
- The molar mass of H (Hydrogen) = 1 g/mol

Thus, the molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

Since NaOH provides one hydroxide ion (OH⁻) per molecule, its n-factor (number of equivalents) is 1. Therefore, the equivalent weight of NaOH is:
\[ E = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{40 \, \text{g/mol}}{1} = 40 \, \text{g/equiv} \]

### Step 4: Calculate the weight of NaOH using the normality.
Now we can rearrange the normality formula to find the weight of NaOH:
\[ W = N \times E \times V \]
Substituting the known values:
- \( N = 0.1 \, \text{N} \)
- \( E = 40 \, \text{g/equiv} \)
- \( V = 0.1 \, \text{L} \)

Calculating:
\[ W = 0.1 \times 40 \times 0.1 = 0.4 \, \text{g} \]

### Conclusion
The weight of NaOH equivalent to 100 ml of 0.1 N oxalic acid is **0.4 grams**.

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