Question

How many g of dibasic acid (mol. wt. 200) should be present in 100 mL of the aqueous solution to give 0.1 normality ?

• A. 1 g
• B. 2 g
• C. 10 g
• D. 20 g

Answer 1

The Correct Option is A

To find out how many grams of dibasic acid (molecular weight 200 g/mol) should be present in 100 mL of solution to achieve 0.1 normality, let's follow these steps:

1. First, understand the concept of normality. For a dibasic acid (acid with two replaceable hydrogen ions per molecule), the normality (N) is calculated using the formula: N = \frac{\text{moles of equivalent solute}}{\text{Volume of solution in liters}}
2. For a dibasic acid, the equivalent weight is half of its molecular weight because it can donate two protons (H₂ equivalents). Therefore, the equivalent weight of the given acid is: = \frac{200}{2} = 100 \, \text{g/equiv}
3. The formula for normality is also expressed as: N = \frac{\text{mass of solute (in grams)}}{\text{equivalent weight} \times \text{Volume of solution (in liters)}}
4. Here, we need 0.1 N in 100 mL of solution. Convert 100 mL to liters, which is 0.1 L.
5. Substitute the values into the normality formula: 0.1 = \frac{\text{mass of solute (in grams)}}{100 \times 0.1}
6. Solving this, we get: \text{mass of solute} = 0.1 \times 100 \times 0.1 = 1 \, \text{g}

Hence, the correct answer is that 1 g of the dibasic acid should be present in the 100 mL solution to achieve 0.1 normality. Therefore, the correct option is 1 g.