0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The molecular weight of the acid will be

Options

A. 45
B. 90
C. 64
D. 128

Explanation

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

### Step 1: Understand the given information
We have:
- Mass of dibasic acid (m) = 0.16 g
- Volume of decinormal NaOH solution (V) = 25 ml
- Decinormal NaOH means Normality (N) = 0.1 N

### Step 2: Calculate the equivalents of NaOH used
Normality (N) is defined as the number of equivalents of solute per liter of solution. Therefore, we can calculate the number of equivalents of NaOH in 25 ml:

\[ \text{Equivalents of NaOH} = N \times \frac{V}{1000} = 0.1 \times \frac{25}{1000} = 0.0025 \text{ equivalents} \]

### Step 3: Relate the equivalents of acid to the equivalents of base
Since the acid is dibasic, it can donate 2 protons (H⁺ ions). Therefore, the number of equivalents of the dibasic acid will be half the number of moles of the acid:

\[ \text{Equivalents of acid} = \frac{\text{mass of acid}}{\text{molecular weight}} \times \text{number of protons donated} \]

For a dibasic acid, the number of protons donated is 2, so:

\[ \text{Equivalents of acid} = \frac{0.16}{\text{Molecular Weight}} \times 2 \]

### Step 4: Set the equivalents of acid equal to the equivalents of NaOH
From the previous steps, we can set the equivalents of acid equal to the equivalents of NaOH:

\[ \frac{0.16 \times 2}{\text{Molecular Weight}} = 0.0025 \]

### Step 5: Solve for the molecular weight
Rearranging the equation to solve for the molecular weight (Mw):

\[ \text{Molecular Weight} = \frac{0.16 \times 2}{0.0025} \]

Calculating this gives:

\[ \text{Molecular Weight} = \frac{0.32}{0.0025} = 128 \text{ g/mol} \]

### Final Answer
The molecular weight of the dibasic acid is **128 g/mol**.

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