Molar Conductivity Numerical

🔢 Numerical Problem: 

 The molar conductivity of a 0.01 M solution of acetic acid (CH₃COOH) is found to be 1.65 × 10⁻² S·m²·mol⁻¹. The molar conductivity at infinite dilution (Λ⁰) for acetic acid is 3.90 × 10⁻² S·m²·mol⁻¹. 

Calculate the degree of dissociation (α) of acetic acid in this solution. 

 ✅ Given: Molar conductivity (Λₘ) = 1.65 × 10⁻² S·m²·mol⁻¹ Molar conductivity at infinite dilution (Λ⁰) = 3.90 × 10⁻² S·m²·mol⁻¹ Molarity (C) = 0.01 mol·L⁻¹

 🧠 Formula Used: $\alpha = \frac{\Lambda_m}{\Lambda^0}$ 

Where: = degree of dissociation = molar conductivity at given concentration

= molar conductivity at infinite dilution 

 Solution: 

 $\alpha = \frac{1.65 \times 10^{-2}}{3.90 \times 10^{-2}}$

 $\alpha = \frac{1.65}{3.90} = 0.423$

 📘 Final Answer: 

 The degree of dissociation (α) of acetic acid = 0.423 or 42.3%