To calculate the elevation in boiling point of a 0.1 M acetic acid (CH₃COOH) solution, we will follow these steps:
### Step 1: Calculate the conductivity (κ) of the solution.
The conductivity (κ) can be calculated using the formula:
\[
\kappa = \frac{1}{R} \cdot \frac{L}{A}
\]
where:
- \( R \) is the resistance (100 Ω),
- \( L \) is the distance between the electrodes (2 cm = 0.02 m),
- \( A \) is the cross-sectional area (4 cm² = 4 × 10⁻⁴ m²).
Substituting the values:
\[
\kappa = \frac{1}{100} \cdot \frac{0.02}{4 \times 10^{-4}} = \frac{0.02}{0.04} = 0.5 \, S/m
\]
### Step 2: Calculate the molar conductivity (Λ) of the solution.
The molar conductivity (Λ) is given by:
\[
\Lambda = \frac{\kappa}{C}
\]
where \( C \) is the concentration in mol/m³. For 0.1 M:
\[
C = 0.1 \, mol/L = 0.1 \times 1000 \, mol/m³ = 100 \, mol/m³
\]
Now substituting the values:
\[
\Lambda = \frac{0.5}{100} = 0.005 \, S \cdot m²/mol
\]
### Step 3: Calculate the degree of dissociation (α).
The molar conductivity of the ions can be calculated using the formula:
\[
\Lambda = \Lambda^0_{H^+} + \Lambda^0_{CH_3COO^-}
\]
Substituting the given values:
\[
\Lambda^0_{H^+} = 300 \, S \cdot cm²/mol = 300 \times 10^{-4} \, S \cdot m²/mol = 0.03 \, S \cdot m²/mol
\]
\[
\Lambda^0_{CH_3COO^-} = 100 \, S \cdot cm²/mol = 100 \times 10^{-4} \, S \cdot m²/mol = 0.01 \, S \cdot m²/mol
\]
Thus,
\[
\Lambda^0 = 0.03 + 0.01 = 0.04 \, S \cdot m²/mol
\]
Now, using the degree of dissociation:
\[
\Lambda = \alpha \Lambda^0
\]
Substituting the values:
\[
0.005 = \alpha \times 0.04
\]
\[
\alpha = \frac{0.005}{0.04} = 0.125
\]
### Step 4: Calculate the elevation in boiling point (ΔT_b).
The formula for elevation in boiling point is:
\[
\Delta T_b = i \cdot K_b \cdot m
\]
where:
- \( i \) is the van 't Hoff factor (for acetic acid, \( i = 1 + \alpha = 1 + 0.125 = 1.125 \)),
- \( K_b = 0.5 \, K \cdot kg/mol \),
- \( m \) is the molality of the solution. For dilute solutions, molality is approximately equal to molarity, so \( m \approx 0.1 \, mol/kg \).
Substituting the values:
\[
\Delta T_b = 1.125 \cdot 0.5 \cdot 0.1 = 0.05625 \, K
\]
### Step 5: Multiply the result by 160.
Finally, we multiply the elevation in boiling point by 160:
\[
\Delta T_b \times 160 = 0.05625 \times 160 = 9 \, K
\]
### Final Answer:
The elevation in boiling point of the 0.1 M acetic acid solution is **9 K**.
---
