An aqueous solution containing `6.5` g of NaCl of `90%` purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is

Options

A. 2000 `cm^(3)`
B. 100 `cm^(3)`
C. 200 `cm^(3)`
D. 1000 `cm^(3)`

Answer

To solve the problem step by step, we need to follow these steps:

### Step 1: Calculate the mass of pure NaCl
Given that the solution contains 6.5 g of NaCl with 90% purity, we can calculate the mass of pure NaCl as follows:

\[ \text{Mass of pure NaCl} = \text{Total mass} \times \text{Purity} = 6.5 \, \text{g} \times 0.9 = 5.85 \, \text{g} \]

### Step 2: Calculate the number of moles of NaCl
The molar mass of NaCl is approximately 58.5 g/mol. We can calculate the number of moles of NaCl using the formula:

\[ \text{Number of moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.1 \, \text{mol} \]

### Step 3: Determine the number of equivalents of NaCl
Since NaCl dissociates completely in solution, the number of equivalents of NaCl is equal to the number of moles:

\[ \text{Number of equivalents of NaCl} = \text{Number of moles of NaCl} = 0.1 \]

### Step 4: Calculate the number of equivalents of NaOH produced
During the electrolysis of NaCl, NaOH is produced. The number of equivalents of NaOH produced will be equal to the number of equivalents of NaCl:

\[ \text{Number of equivalents of NaOH} = 0.1 \]

### Step 5: Calculate the volume of 1 M acetic acid required for neutralization
The neutralization reaction between NaOH and acetic acid (CH₃COOH) is a 1:1 reaction. Therefore, the number of equivalents of acetic acid required will also be equal to the number of equivalents of NaOH:

\[ \text{Volume of 1 M acetic acid required} = \text{Number of equivalents of acetic acid} = 0.1 \, \text{L} = 100 \, \text{mL} \]

### Final Answer
The volume of 1 M acetic acid required to neutralize the NaOH obtained is **100 mL**.

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