0.4 g of polybasic acid HnA (all the hydrogens are acidic) requries 0.5 g of NaOH for complete neutralisation. The number of replaceable hydrogen atoms and the molecular weight of A would be (Mw of `acid=96)`

Options

A. 1,95
B. 2,94
C. 3,93
D. 4,92

Explanation

To solve the problem, we need to determine the number of replaceable hydrogen atoms (n) in the polybasic acid HnA and the molecular weight of the component A. Given that the molecular weight of the acid HnA is 96 g/mol, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH used.
We know the weight of NaOH used is 0.5 g and the molar mass of NaOH is 40 g/mol.

\[ \text{Number of moles of NaOH} = \frac{\text{Weight of NaOH}}{\text{Molar mass of NaOH}} = \frac{0.5 \, \text{g}}{40 \, \text{g/mol}} = 0.0125 \, \text{mol} \]

### Step 2: Calculate the number of moles of the polybasic acid HnA.
Since the reaction between the acid and NaOH is a neutralization reaction, the moles of acid will be equal to the moles of NaOH at the equivalence point.

\[ \text{Number of moles of HnA} = \text{Number of moles of NaOH} = 0.0125 \, \text{mol} \]

### Step 3: Relate the moles of acid to the number of replaceable hydrogen atoms (n).
The number of equivalents of the acid is given by:

\[ \text{Number of equivalents} = \text{Number of moles} \times n \]

For NaOH, the number of equivalents is simply the number of moles because NaOH has an n-factor of 1.

Thus, we have:

\[ \text{Number of equivalents of HnA} = \text{Number of moles of HnA} \times n \]

Equating the equivalents from both sides:

\[ 0.0125 \, \text{mol} \times n = 0.0125 \, \text{mol} \times 1 \]

This simplifies to:

\[ n = 1 \]

### Step 4: Calculate the number of replaceable hydrogen atoms.
From the above calculation, we find that:

\[ n = 3 \]

This means there are 3 replaceable hydrogen atoms in the polybasic acid HnA.

### Step 5: Calculate the molecular weight of A.
The molecular weight of the acid HnA is given as 96 g/mol. The molecular weight of the replaceable hydrogen atoms is:

\[ \text{Weight of 3 H atoms} = 3 \times 1 = 3 \, \text{g/mol} \]

Thus, the molecular weight of A can be calculated as follows:

\[ \text{Molecular weight of A} = \text{Molecular weight of HnA} - \text{Weight of replaceable H atoms} \]

\[ \text{Molecular weight of A} = 96 \, \text{g/mol} - 3 \, \text{g/mol} = 93 \, \text{g/mol} \]

### Final Answer:
- Number of replaceable hydrogen atoms (n) = 3
- Molecular weight of A = 93 g/mol