Calculate the normality of solution containing 31.5 g of hydrated oxalic acid `(H_(2)C_(2)O_(4),2H_(2)O)` in 1250 mL of solution. Multiply answer with 10

Explanation

To calculate the normality of the solution containing 31.5 g of hydrated oxalic acid \((H_2C_2O_4 \cdot 2H_2O)\) in 1250 mL of solution, we will follow these steps:

### Step 1: Determine the Molar Mass of Hydrated Oxalic Acid
The formula for hydrated oxalic acid is \(H_2C_2O_4 \cdot 2H_2O\). We need to calculate its molar mass.

- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

Calculating the molar mass:
- For \(H_2C_2O_4\):
- \(2 \times 12\) (for C) = 24 g
- \(4 \times 16\) (for O) = 64 g
- \(2 \times 1\) (for H) = 2 g
- Total for \(H_2C_2O_4\) = \(24 + 64 + 2 = 90\) g/mol

- For \(2H_2O\):
- \(2 \times (2 \times 1 + 16)\) = \(2 \times 18\) = 36 g/mol

Total molar mass of \(H_2C_2O_4 \cdot 2H_2O = 90 + 36 = 126\) g/mol.

### Step 2: Calculate the Equivalent Weight
The equivalent weight of an acid is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
where \(n\) is the number of dissociable hydrogen ions.

For oxalic acid, \(n = 2\) (since it can donate 2 protons).

So, the equivalent weight of hydrated oxalic acid is:
\[ \text{Equivalent Weight} = \frac{126 \text{ g/mol}}{2} = 63 \text{ g/equiv} \]

### Step 3: Calculate the Number of Gram Equivalents
The number of gram equivalents can be calculated using the formula:
\[ \text{Number of Gram Equivalents} = \frac{\text{Given Mass}}{\text{Equivalent Weight}} \]
Given mass = 31.5 g.

So,
\[ \text{Number of Gram Equivalents} = \frac{31.5 \text{ g}}{63 \text{ g/equiv}} = 0.5 \text{ equiv} \]

### Step 4: Calculate the Volume in Liters
Convert the volume from mL to L:
\[ 1250 \text{ mL} = \frac{1250}{1000} = 1.25 \text{ L} \]

### Step 5: Calculate the Normality
Normality (N) is given by:
\[ N = \frac{\text{Number of Gram Equivalents}}{\text{Volume of Solution in Liters}} \]
Substituting the values:
\[ N = \frac{0.5 \text{ equiv}}{1.25 \text{ L}} = 0.4 \text{ N} \]

### Step 6: Multiply the Normality by 10
Finally, we multiply the normality by 10:
\[ 0.4 \times 10 = 4.0 \]

### Final Answer
The final answer is \(4.0\).

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