How many grams of dibasic acid (mol. weight 100) should be present in 200 mL of the aqueous solution to give strength 0.3 N?

Options

A. 1g
B. 2g
C. 3g
D. 4g

Explanation

To solve the problem of how many grams of dibasic acid (molecular weight = 100 g/mol) should be present in 200 mL of aqueous solution to give a strength of 0.3 N, we can follow these steps:

### Step 1: Understand Normality
Normality (N) is defined as the number of gram equivalents of solute per liter of solution. The formula for normality is:
\[ N = \frac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in liters}} \]

### Step 2: Calculate the Volume in Liters
Given that the volume of the solution is 200 mL, we need to convert this to liters:
\[ \text{Volume in liters} = \frac{200 \, \text{mL}}{1000} = 0.2 \, \text{L} \]

### Step 3: Calculate the Equivalent Weight
The equivalent weight of a dibasic acid can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar weight}}{\text{Basicity}} \]
For a dibasic acid, the basicity is 2. Therefore:
\[ \text{Equivalent weight} = \frac{100 \, \text{g/mol}}{2} = 50 \, \text{g/equiv} \]

### Step 4: Set Up the Normality Equation
We can rearrange the normality equation to find the mass of the solute:
\[ N = \frac{\text{mass}}{\text{Equivalent weight} \times \text{Volume in liters}} \]
Substituting the known values:
\[ 0.3 = \frac{\text{mass}}{50 \times 0.2} \]

### Step 5: Solve for Mass
Now we can solve for mass:
\[ 0.3 = \frac{\text{mass}}{10} \]
\[ \text{mass} = 0.3 \times 10 = 3 \, \text{grams} \]

### Conclusion
The mass of dibasic acid required to achieve a strength of 0.3 N in 200 mL of solution is **3 grams**.

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