A bottle, which contains 200 ml of 0.1 M KOH, absorbs 1 millimole of `CO_(2)` from the air. If the solution is then treated with standard acid using phenolphthalein indicator, the normality of solution formed by absorption of `CO_(2)` is :

Options

A. `0.095N`
B. 0.1N
C. 0.2N
D. 0.036N

Explanation

To solve the problem step-by-step, we will follow the outlined procedure:

### Step 1: Calculate the number of millimoles of KOH in the solution.
Given:
- Volume of KOH solution = 200 ml = 0.2 L
- Molarity of KOH = 0.1 M

Using the formula:
\[ \text{Millimoles of KOH} = \text{Molarity} \times \text{Volume in L} \times 1000 \]
\[ \text{Millimoles of KOH} = 0.1 \times 0.2 \times 1000 = 20 \text{ millimoles} \]

### Step 2: Determine the reaction between KOH and CO2.
The reaction is:
\[ 2 \text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]
From the stoichiometry of the reaction, 2 millimoles of KOH react with 1 millimole of CO2.

### Step 3: Calculate the amount of KOH that reacts with CO2.
Given that 1 millimole of CO2 is absorbed, the amount of KOH that reacts is:
\[ \text{KOH reacted} = 2 \times 1 = 2 \text{ millimoles} \]

### Step 4: Calculate the unreacted KOH.
\[ \text{Unreacted KOH} = \text{Initial KOH} - \text{KOH reacted} \]
\[ \text{Unreacted KOH} = 20 - 2 = 18 \text{ millimoles} \]

### Step 5: Determine the amount of K2CO3 produced.
From the reaction, 1 millimole of CO2 produces 1 millimole of K2CO3. Therefore:
\[ \text{K}_2\text{CO}_3 \text{ produced} = 1 \text{ millimole} \]

### Step 6: Calculate the total equivalents of KOH and K2CO3.
- KOH contributes 18 milliequivalents (since 1 millimole of KOH = 1 milliequivalent).
- K2CO3 contributes 1 milliequivalent (since K2CO3 can donate 1 equivalent of K+ ions).

Total milliequivalents:
\[ \text{Total milliequivalents} = \text{Unreacted KOH} + \text{K}_2\text{CO}_3 = 18 + 1 = 19 \text{ milliequivalents} \]

### Step 7: Calculate the normality of the solution.
Normality (N) is defined as:
\[ N = \frac{\text{Number of equivalents}}{\text{Volume of solution in L}} \]
The volume of the solution is 200 ml = 0.2 L.

\[ N = \frac{19 \text{ milliequivalents}}{0.2 \text{ L}} = 0.095 \text{ N} \]

### Final Answer:
The normality of the solution formed by the absorption of CO2 is **0.095 N**.

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