Question

The molar conductivity of \(CH_3COOH\) at a certain concentration is \(39.0\ \Omega^{-1}\,cm^2\,mol^{-1}\). Given that the limiting molar conductivities are: \[ \Lambda_m^\circ(HCl)=426.0\ \Omega^{-1}\,cm^2\,mol^{-1} \] \[ \Lambda_m^\circ(CH_3COONa)=91.0\ \Omega^{-1}\,cm^2\,mol^{-1} \] \[ \Lambda_m^\circ(NaCl)=126.0\ \Omega^{-1}\,cm^2\,mol^{-1} \] The degree of dissociation of acetic acid at this concentration is:

• A. \(0.25\)
• B. \(0.40\)
• C. \(0.50\)
• D. \(0.60\)

Hint:

For weak electrolytes: \[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \] Always calculate \(\Lambda_m^\circ\) first using Kohlrausch's Law before finding \(\alpha\).

Answer 1

The Correct Option is C

Concept: This question combines:
• Kohlrausch's Law
• Limiting molar conductivity
• Degree of dissociation
• Weak electrolyte behaviour Such integrated questions are very common in CUET because they test conceptual understanding rather than direct memorization. For weak electrolytes: \[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \] where: \[ \alpha=\text{Degree of dissociation} \] \[ \Lambda_m=\text{Molar conductivity at given concentration} \] \[ \Lambda_m^\circ=\text{Limiting molar conductivity} \]

Step 1: Calculate the limiting molar conductivity of acetic acid. Using Kohlrausch's Law: \[ \Lambda_m^\circ(CH_3COOH) = \Lambda_m^\circ(HCl) + \Lambda_m^\circ(CH_3COONa) - \Lambda_m^\circ(NaCl) \] Substituting values: \[ = 426+91-126 \] \[ = 391 \] Therefore: \[ \Lambda_m^\circ(CH_3COOH) = 391\times10^{-1} = 78.0 \] \[ \boxed{\Lambda_m^\circ=78.0\ \Omega^{-1}cm^2mol^{-1}} \]

Step 2: Apply the degree of dissociation formula. Given: \[ \Lambda_m=39.0 \] Thus: \[ \alpha = \frac{39.0}{78.0} \] \[ = 0.50 \]

Step 3: Interpret the result. This means: \[ 50% \] of the acetic acid molecules are ionized at the given concentration. The remaining: \[ 50% \] remain unionized.

Step 4: Verification. \[ 78\times0.50 = 39 \] Hence the answer is correct.

Step 5: Final conclusion. \[ \boxed{\alpha=0.50} \] Therefore: \[ \boxed{\text{Option (C)}} \]