Step 1: Calculate the moles of acetic acid in the sample
Given:
- Mass of acetic acid (\( \text{CH}_3\text{COOH} \)) = 60 g
- Mole fraction of acetic acid = 0.2
Let's first find the moles of acetic acid. We need the molar mass of acetic acid to do this.
Molar mass of acetic acid (CH\(_3\)COOH):
- Carbon (C): \( 12.01 \, \text{g/mol} \)
- Hydrogen (H): \( 1.01 \, \text{g/mol} \)
- Oxygen (O): \( 16.00 \, \text{g/mol} \)
\[ \text{Molar mass of CH}_3\text{COOH} = 2(12.01) + 4(1.01) + 2(16.00) = 24.02 + 4.04 + 32.00 = 60.06 \, \text{g/mol} \]
Calculate the moles of acetic acid:
\[ \text{Moles of CH}_3\text{COOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \, \text{g}}{60.06 \, \text{g/mol}} \approx 0.999 \, \text{mol} \]
Step 2: Determine the moles of KOH required for neutralization
The balanced chemical equation for the neutralization reaction between acetic acid and KOH is:
\[ \text{CH}_3\text{COOH} + \text{KOH} \rightarrow \text{CH}_3\text{COOK} + \text{H}_2\text{O} \]
From the equation, we see that 1 mole of acetic acid reacts with 1 mole of KOH. Therefore, the moles of KOH required are equal to the moles of acetic acid.
\[ \text{Moles of KOH required} = 0.999 \, \text{mol} \]
Step 3: Calculate the volume of 0.02 M KOH solution needed
Given:
- Molarity of KOH solution = 0.02 M
- Moles of KOH required = 0.999 mol
Calculate the volume of KOH solution:
\[ \text{Volume (L)} = \frac{\text{Moles of solute}}{\text{Molarity}} = \frac{0.999 \, \text{mol}}{0.02 \, \text{M}} = 49.95 \, \text{L} \]
So, the volume of 0.02 M KOH solution required to neutralize 60 g of acetic acid is approximately 49.95 liters.
