Calculate the weight of sodium oxalate present in 250 mL of seminormal solution.

Answer

To solve the problem of calculating the weight of sodium oxalate present in 250 mL of a semi-normal solution, we can follow these steps:

### Step 1: Determine the formula and molar mass of sodium oxalate.
The chemical formula for sodium oxalate is \( \text{Na}_2\text{C}_2\text{O}_4 \).

**Calculation of Molar Mass:**
- Sodium (Na): 2 atoms × 23 g/mol = 46 g/mol
- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Oxygen (O): 4 atoms × 16 g/mol = 64 g/mol

Adding these together:
\[ \text{Molar Mass} = 46 + 24 + 64 = 134 \text{ g/mol} \]

### Step 2: Calculate the equivalent mass of sodium oxalate.
The equivalent mass can be calculated using the formula:
\[ \text{Equivalent Mass} = \frac{\text{Molar Mass}}{n} \]
where \( n \) is the number of equivalents. For sodium oxalate, it can donate 2 moles of sodium ions (Na\(^+\)), so \( n = 2 \).

Thus,
\[ \text{Equivalent Mass} = \frac{134 \text{ g/mol}}{2} = 67 \text{ g/equiv} \]

### Step 3: Calculate the number of equivalents in the solution.
We know that normality (N) is defined as the number of equivalents per liter of solution. A semi-normal solution means it has a normality of \( \frac{1}{2} \) N or 0.5 N.

Given that we have 250 mL of solution, we convert this to liters:
\[ \text{Volume in Liters} = \frac{250 \text{ mL}}{1000} = 0.25 \text{ L} \]

Now, we can calculate the number of equivalents:
\[ \text{Number of Equivalents} = \text{Normality} \times \text{Volume in Liters} = 0.5 \text{ N} \times 0.25 \text{ L} = 0.125 \text{ equivalents} \]

### Step 4: Calculate the weight of sodium oxalate.
Using the number of equivalents and the equivalent mass, we can find the weight of sodium oxalate:
\[ \text{Weight (W)} = \text{Number of Equivalents} \times \text{Equivalent Mass} \]
\[ W = 0.125 \text{ equivalents} \times 67 \text{ g/equiv} = 8.375 \text{ g} \]

### Final Answer:
The weight of sodium oxalate present in 250 mL of a semi-normal solution is **8.375 grams**.

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