To find the basicity of the acid, we will follow these steps:
### Step 1: Calculate the molarity of the acid solution.
The acid has a concentration of 29.4 g/L. To find the molarity (M), we use the formula:
\[
\text{Molarity (M)} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}}
\]
Given:
- Mass of acid = 29.4 g
- Molar mass of acid = 98 g/mol
- Volume of solution = 1 L
\[
\text{Molarity} = \frac{29.4 \, \text{g}}{98 \, \text{g/mol} \times 1 \, \text{L}} = 0.3 \, \text{mol/L}
\]
### Step 2: Calculate the number of moles of acid in 100 cm³ of the solution.
To find the number of moles of acid in 100 cm³ (0.1 L):
\[
\text{Number of moles of acid} = \text{Molarity} \times \text{Volume (L)}
\]
\[
\text{Number of moles of acid} = 0.3 \, \text{mol/L} \times 0.1 \, \text{L} = 0.03 \, \text{mol}
\]
### Step 3: Calculate the molarity of the NaOH solution.
The NaOH solution contains 20 g of NaOH per 500 cm³. First, we need to find the number of moles of NaOH:
\[
\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}
\]
\[
\text{Number of moles of NaOH} = \frac{20 \, \text{g}}{40 \, \text{g/mol}} = 0.5 \, \text{mol}
\]
Now, to find the molarity of the NaOH solution:
\[
\text{Molarity of NaOH} = \frac{0.5 \, \text{mol}}{0.5 \, \text{L}} = 1 \, \text{mol/L}
\]
### Step 4: Calculate the number of moles of NaOH in 90 cm³.
To find the number of moles of NaOH in 90 cm³ (0.09 L):
\[
\text{Number of moles of NaOH} = \text{Molarity} \times \text{Volume (L)}
\]
\[
\text{Number of moles of NaOH} = 1 \, \text{mol/L} \times 0.09 \, \text{L} = 0.09 \, \text{mol}
\]
### Step 5: Set up the neutralization reaction.
The neutralization reaction can be represented as:
\[
\text{Acid} + \text{NaOH} \rightarrow \text{Salt} + \text{Water}
\]
Let the basicity of the acid be \( n \). The reaction shows that \( n \) moles of the acid react with 1 mole of NaOH.
### Step 6: Relate the moles of acid and NaOH.
From the stoichiometry of the reaction, we have:
\[
n \times \text{moles of acid} = \text{moles of NaOH}
\]
Substituting the values we calculated:
\[
n \times 0.03 \, \text{mol} = 0.09 \, \text{mol}
\]
### Step 7: Solve for \( n \).
\[
n = \frac{0.09 \, \text{mol}}{0.03 \, \text{mol}} = 3
\]
### Conclusion:
The basicity of the acid is 3.
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