`12.5 mL` of a solution containing `6.0 g` of a dibasic acid in `1 L` was found to be neutralized by `10 mL` a decinormal solution of `NaOH`. The molecular weight of the acid is

Options

A. 150
B. 120
C. 110
D. 75

Explanation

To find the molecular weight of the dibasic acid, we can follow these steps:

### Step 1: Understand the given information
We have:
- Volume of the acid solution (V1) = 12.5 mL = 0.0125 L
- Mass of the dibasic acid = 6.0 g
- Volume of NaOH solution (V2) = 10 mL = 0.010 L
- Normality of NaOH solution (N2) = 0.1 N (decinormal)

### Step 2: Use the neutralization equation
The neutralization reaction can be expressed using the formula:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) = Normality of the acid
- \( V_1 \) = Volume of the acid solution
- \( N_2 \) = Normality of NaOH
- \( V_2 \) = Volume of NaOH solution

### Step 3: Substitute the known values
Substituting the values into the equation:
\[ N_1 \times 0.0125 = 0.1 \times 0.010 \]

### Step 4: Calculate \( N_1 \)
Rearranging the equation to find \( N_1 \):
\[ N_1 = \frac{0.1 \times 0.010}{0.0125} \]
\[ N_1 = \frac{0.001}{0.0125} \]
\[ N_1 = 0.08 \, \text{N} \]

### Step 5: Calculate the equivalent weight of the acid
The strength of the acid solution can be calculated using the formula:
\[ \text{Strength} = N_1 \times \text{Equivalent Weight} \]
Given that the strength of the acid solution is 6.0 g in 1 L, we can write:
\[ 6.0 = 0.08 \times \text{Equivalent Weight} \]

### Step 6: Solve for the equivalent weight
Rearranging the equation:
\[ \text{Equivalent Weight} = \frac{6.0}{0.08} \]
\[ \text{Equivalent Weight} = 75 \, \text{g} \]

### Step 7: Calculate the molecular weight of the dibasic acid
Since the acid is dibasic, its basicity is 2. The molecular weight (M) can be calculated as:
\[ M = \text{Equivalent Weight} \times \text{Basicity} \]
\[ M = 75 \times 2 \]
\[ M = 150 \, \text{g/mol} \]

### Conclusion
The molecular weight of the dibasic acid is **150 g/mol**.

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