0.84 g of a acid (mol wt. 150) was dissolved in water and the volume was made up to 100 ml. 25 ml of this solution required 28 ml of (N/10) NaOH solution for neutralisation. The equivalent weight and basicity of the acid

Options

A. 75,2
B. 150,1
C. 75,4
D. 150,2

Answer

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Calculate the number of moles of the acid
Given:
- Mass of the acid = 0.84 g
- Molar mass of the acid = 150 g/mol

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.84 \, \text{g}}{150 \, \text{g/mol}} = 0.0056 \, \text{mol} \]

### Step 2: Calculate the number of equivalents of the acid
Let \( n \) be the basicity (or n-factor) of the acid. The number of equivalents of the acid can be expressed as:
\[ \text{Number of equivalents} = \text{Number of moles} \times n = 0.0056 \times n \]

### Step 3: Calculate the normality of the acid solution
The total volume of the solution is 100 ml. Therefore, the normality \( N \) of the acid can be calculated using the formula:
\[ N = \frac{\text{Number of equivalents}}{\text{Volume of solution in L}} = \frac{0.0056 \times n}{0.1} = 0.056n \, \text{N} \]

### Step 4: Use the neutralization reaction to find \( n \)
From the problem, we know that:
- Volume of acid solution used = 25 ml
- Volume of NaOH solution used = 28 ml
- Normality of NaOH solution = \( \frac{N}{10} = 0.1 \, \text{N} \)

Using the equation for neutralization:
\[ V_1 \times N_1 = V_2 \times N_2 \]
Substituting the known values:
\[ 25 \times (0.056n) = 28 \times 0.1 \]
\[ 1.4n = 2.8 \]
\[ n = \frac{2.8}{1.4} = 2 \]

### Step 5: Calculate the equivalent weight of the acid
The equivalent weight of the acid can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} = \frac{150}{2} = 75 \, \text{g/equiv} \]

### Final Answer
- Equivalent weight of the acid = 75 g/equiv
- Basicity of the acid = 2

### Summary
The equivalent weight of the acid is 75 g/equiv and its basicity is 2.

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