`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`

Explanation

To find the basicity of the given acid, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH used.
The formula for calculating the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]
Given:
- Molarity of NaOH = 0.5 M
- Volume of NaOH = 20 mL = 0.020 L

Calculating the number of moles of NaOH:
\[ \text{Number of moles of NaOH} = 0.5 \, \text{mol/L} \times 0.020 \, \text{L} = 0.01 \, \text{mol} \]

### Step 2: Calculate the equivalent moles of NaOH.
Since NaOH is a strong base with a basicity of 1, the number of equivalent moles of NaOH is the same as the number of moles:
\[ \text{Equivalent moles of NaOH} = 0.01 \, \text{eq} \]

### Step 3: Calculate the number of gram-equivalent moles of the acid.
The acid is neutralized by NaOH, which means the equivalent moles of the acid must equal the equivalent moles of NaOH:
\[ \text{Equivalent moles of acid} = \text{Equivalent moles of NaOH} = 0.01 \, \text{eq} \]

### Step 4: Calculate the number of gram-equivalent moles of the acid using its weight and molecular weight.
The formula for equivalent moles is:
\[ \text{Equivalent moles} = \frac{\text{Weight (g)}}{\text{Molecular Weight (g/mol)} \times \text{Basicity}} \]
Given:
- Weight of the acid = 0.45 g
- Molecular Weight of the acid = 90 g/mol

Substituting the values:
\[ 0.01 = \frac{0.45}{90 \times \text{Basicity}} \]

### Step 5: Solve for the basicity of the acid.
Rearranging the equation to solve for Basicity:
\[ \text{Basicity} = \frac{0.45}{90 \times 0.01} = \frac{0.45}{0.9} = 0.5 \]

### Step 6: Final Calculation
Since we need the basicity in whole numbers, we can multiply by 2 to convert it to a whole number:
\[ \text{Basicity} = 2 \]

### Conclusion
The basicity of the given acid is **2**.

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