How many grams of a dibasic acid (Mol. Wt. =200) should be present in 100 ml of its aqueous solution to give decinormal strength

Answer

To solve the problem of how many grams of a dibasic acid (with a molecular weight of 200 g/mol) should be present in 100 ml of its aqueous solution to give a decinormal strength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Normality**:
- Normality (N) is defined as the number of equivalents of solute per liter of solution. For a dibasic acid, which can donate two protons (H⁺ ions), the normality will be twice the molarity.

2. **Identify the Given Values**:
- Molecular weight of the dibasic acid = 200 g/mol
- Volume of solution = 100 ml = 0.1 L (since 1000 ml = 1 L)
- Desired normality = 0.1 N (decinormal strength)

3. **Calculate the Equivalent Weight**:
- For a dibasic acid, the equivalent weight is calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Number of acidic protons}} = \frac{200 \, \text{g/mol}}{2} = 100 \, \text{g/equiv} \]

4. **Use the Normality Formula**:
- The formula for normality is:
\[ N = \frac{\text{Weight of solute (g)}}{\text{Equivalent weight (g/equiv)} \times \text{Volume of solution (L)}} \]
- Rearranging the formula to find the weight of the solute gives:
\[ \text{Weight of solute (g)} = N \times \text{Equivalent weight (g/equiv)} \times \text{Volume of solution (L)} \]

5. **Substitute the Values**:
- Substitute the known values into the equation:
\[ \text{Weight of solute} = 0.1 \, \text{N} \times 100 \, \text{g/equiv} \times 0.1 \, \text{L} \]
- Calculate:
\[ \text{Weight of solute} = 0.1 \times 100 \times 0.1 = 1 \, \text{g} \]

6. **Conclusion**:
- Therefore, to prepare a decinormal solution of the dibasic acid, you need **1 gram** of the acid in 100 ml of aqueous solution.