With 63 gm of oxalic acid how many litres of `(N)/(10)` solution can be prepared

Options

A. 100 litre
B. 10 litre
C. 1 litre
D. 1000 litre

Answer

To solve the problem of how many liters of a \( \frac{N}{10} \) (0.1 N) solution can be prepared with 63 g of oxalic acid, we will follow these steps:

### Step 1: Determine the Molar Mass of Oxalic Acid
The formula for oxalic acid is \( C_2H_2O_4 \cdot 2H_2O \) (or \( H_2C_2O_4 \cdot 2H_2O \)).
- The molar mass of oxalic acid is calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 2 + 2 g/mol × 2 = 4 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol
- Water (H2O): 18 g/mol × 2 = 36 g/mol
- Total molar mass = 24 + 4 + 64 + 36 = 128 g/mol

### Step 2: Calculate the Number of Moles of Oxalic Acid
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{63 \, \text{g}}{126 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 3: Determine the Equivalent Weight of Oxalic Acid
The equivalent weight is calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
where \( n \) is the number of hydrogen ions that can be replaced or the number of equivalents. For oxalic acid, \( n = 2 \) (as it can donate two protons).
\[ \text{Equivalent weight} = \frac{126 \, \text{g/mol}}{2} = 63 \, \text{g/equiv} \]

### Step 4: Calculate the Number of Gram Equivalents
Using the formula:
\[ \text{Number of gram equivalents} = \frac{\text{Weight of solute}}{\text{Equivalent weight}} \]
Substituting the values:
\[ \text{Number of gram equivalents} = \frac{63 \, \text{g}}{63 \, \text{g/equiv}} = 1 \, \text{equiv} \]

### Step 5: Use Normality to Find Volume
Normality (N) is defined as:
\[ N = \frac{\text{Number of gram equivalents}}{\text{Volume in liters}} \]
Given that the normality is \( 0.1 \, N \):
\[ 0.1 = \frac{1 \, \text{equiv}}{V} \]
Rearranging to find the volume \( V \):
\[ V = \frac{1 \, \text{equiv}}{0.1} = 10 \, \text{liters} \]

### Final Answer
With 63 g of oxalic acid, you can prepare **10 liters** of a \( \frac{N}{10} \) solution.

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