Question

The limiting molar conductivities of \( HCl \), \( CH_3COONa \), and \( NaCl \) are respectively 425, 90, and 125 mho cm\(^2\) mol\(^{-1}\) at 25°C. The molar conductivity of 0.1M \( CH_3COOH \) solution is 7.8 mho cm\(^2\) mol\(^{-1}\) at the same temperature. The degree of dissociation of 0.1M acetic acid solution at the same temperature is:

• A. \( 0.10 \)
• B. \( 0.02 \)
• C. \( 0.15 \)
• D. \( 0.03 \)

Hint:

Kohlrausch’s law states: \[ \lambda_m^\infty ({Weak Electrolyte}) = \lambda_m^\infty ({Strong Acid}) + \lambda_m^\infty ({Salt}) - \lambda_m^\infty ({Common Ion Salt}) \]

Answer 1

The Correct Option is B

Step 1: Finding \( \lambda_m^\infty \) for acetic acid.
Using Kohlrausch’s law: \[ \lambda_m^\infty (CH_3COOH) = \lambda_m^\infty (HCl) + \lambda_m^\infty (CH_3COONa) - \lambda_m^\infty (NaCl) \] \[ = 425 + 90 - 125 = 390 { mho cm}^2 { mol}^{-1} \] Step 2: Finding the degree of dissociation (\( \alpha \)).
\[ \alpha = \frac{\lambda_m}{\lambda_m^\infty} \] \[ \alpha = \frac{7.8}{390} = 0.02 \]