`0.45 g` of acid (mol. Wt.`=90`) was exactly neutralized by `20 ml` of `0.5(M) NaOH`.
The basicity of the given acid is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of determining the basicity of the given acid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the acid (m) = 0.45 g
- Molecular weight of the acid (M) = 90 g/mol
- Volume of NaOH solution (V) = 20 ml = 0.020 L (convert ml to L)
- Molarity of NaOH (C) = 0.5 M

2. **Calculate the Number of Moles of NaOH:**
The number of moles of NaOH can be calculated using the formula:
\[ \text{Number of moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Number of moles of NaOH} = 0.5 \, \text{mol/L} \times 0.020 \, \text{L} = 0.01 \, \text{mol} \]

3. **Calculate the Number of Equivalents of NaOH:**
Since NaOH is a strong base and has an n-factor of 1 (it can donate one hydroxide ion, OH⁻), the number of equivalents of NaOH is equal to the number of moles:
\[ \text{Number of equivalents of NaOH} = \text{Number of moles of NaOH} \times \text{n-factor} \]
\[ \text{Number of equivalents of NaOH} = 0.01 \, \text{mol} \times 1 = 0.01 \, \text{equivalents} \]

4. **Calculate the Number of Moles of the Acid:**
The number of moles of the acid can be calculated using the formula:
\[ \text{Number of moles of acid} = \frac{\text{Mass of acid}}{\text{Molecular weight of acid}} \]
\[ \text{Number of moles of acid} = \frac{0.45 \, \text{g}}{90 \, \text{g/mol}} = 0.005 \, \text{mol} \]

5. **Calculate the Number of Equivalents of the Acid:**
The number of equivalents of the acid can be expressed as:
\[ \text{Number of equivalents of acid} = \text{Number of moles of acid} \times \text{n-factor} \]
Since the acid is neutralized by NaOH, the number of equivalents of acid is equal to the number of equivalents of NaOH:
\[ 0.01 \, \text{equivalents of acid} = 0.005 \, \text{mol} \times \text{n-factor} \]
Rearranging gives:
\[ \text{n-factor} = \frac{0.01 \, \text{equivalents}}{0.005 \, \text{mol}} = 2 \]

6. **Conclusion:**
The basicity of the acid, which is equal to its n-factor, is 2. This means the acid can donate 2 protons (H⁺ ions), indicating that it is a dibasic acid.

### Final Answer:
The basicity of the given acid is **2**.