0.45 g of a dibasic acid is completely neutralised with 100 ml `N/10` NaOH . The molecular weight of acid is

Options

A. 45
B. 90
C. 180
D. 22.5

Answer

To find the molecular weight of the dibasic acid that is completely neutralized by NaOH, we can follow these steps:

### Step 1: Understand the Problem
We have a dibasic acid that weighs 0.45 g and is neutralized by 100 ml of `N/10` NaOH. We need to find the molecular weight of the acid.

### Step 2: Calculate the Number of Gram Equivalents of NaOH
The normality (N) of NaOH is given as `N/10`, which is equivalent to 0.1 N. The volume of NaOH used is 100 ml, which we need to convert to liters:
- Volume in liters = 100 ml = 0.1 L

Using the formula for the number of gram equivalents:
\[ \text{Number of gram equivalents of NaOH} = \text{Normality} \times \text{Volume in liters} \]
\[ = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 3: Relate the Gram Equivalents of Acid and NaOH
According to the law of equivalence:
\[ \text{Number of gram equivalents of dibasic acid} = \text{Number of gram equivalents of NaOH} \]
Thus, the number of gram equivalents of the dibasic acid is also 0.01 equivalents.

### Step 4: Calculate the Equivalent Weight of the Acid
The equivalent weight of the dibasic acid can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Weight of acid}}{\text{Number of gram equivalents}} \]
Substituting the values:
\[ \text{Equivalent weight} = \frac{0.45 \, \text{g}}{0.01} = 45 \, \text{g/equiv} \]

### Step 5: Determine the Molecular Weight of the Acid
Since the acid is dibasic, it can donate two protons (H⁺ ions). Therefore, the n-factor (number of acidic protons) is 2. The molecular weight (MW) can be calculated as:
\[ \text{Molecular weight} = \text{Equivalent weight} \times \text{n-factor} \]
Substituting the values:
\[ \text{Molecular weight} = 45 \, \text{g/equiv} \times 2 = 90 \, \text{g/mol} \]

### Conclusion
The molecular weight of the dibasic acid is **90 g/mol**.