The amount of oxalic acid (eq. wt 63) required to preare 500 ml of its 0.10 N solution is

Options

A. `0.315g`
B. `3.150g`
C. `6.300g`
D. `63.00g`

Explanation

To find the amount of oxalic acid required to prepare 500 ml of a 0.10 N solution, we can follow these steps:

### Step 1: Understand the formula for Normality
Normality (N) is defined as the number of equivalents of solute per liter of solution. The formula can be expressed as:
\[ N = \frac{\text{Gram equivalents of solute}}{\text{Volume of solution in liters}} \]

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

### Step 3: Use the formula for Gram Equivalents
The number of gram equivalents can be calculated using the formula:
\[ \text{Gram equivalents} = \frac{\text{Given weight}}{\text{Equivalent weight}} \]
Where the equivalent weight of oxalic acid is given as 63 g.

### Step 4: Rearranging the Normality formula
From the Normality formula, we can rearrange it to find the given weight (x):
\[ x = N \times \text{Equivalent weight} \times \text{Volume in liters} \]

### Step 5: Substitute the known values
Now, substituting the known values into the equation:
- Normality (N) = 0.10 N
- Equivalent weight = 63 g
- Volume in liters = 0.5 L

Thus,
\[ x = 0.10 \times 63 \times 0.5 \]

### Step 6: Calculate the amount of oxalic acid
Now we perform the calculation:
\[ x = 0.10 \times 63 \times 0.5 = 3.15 \text{ g} \]

### Conclusion
The amount of oxalic acid required to prepare 500 ml of a 0.10 N solution is **3.15 g**.

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