Question

The limiting molar conductances of NaCl, HCl and CH\(_3\)COONa at 300 K are 126.4, 425.9 and 91.0 S cm\(^2\) mol\(^-1\) respectively. The limiting molar conductance of acetic acid at 300 K is:

• A. \(266 \, {S cm}^2 \, {mol}^{-1}\)
• B. \(390.5 \, {S cm}^2 \, {mol}^{-1}\)
• C. \(461.3 \, {S cm}^2 \, {mol}^{-1}\)
• D. \(208 \, {S cm}^2 \, {mol}^{-1}\)
• E. \(108 \, {S cm}^2 \, {mol}^{-1}\)

Hint:

The limiting molar conductance is the maximum conductance when the concentration of the electrolyte approaches zero.

Answer 1

The Correct Option is B

We use the formula for the limiting molar conductance of acetic acid: \[ \Lambda_m = \Lambda_m ({NaCl}) + \Lambda_m ({HCl}) - \Lambda_m ({CH}_3{COONa}) \] Substituting the given values: \[ \Lambda_m = 126.4 + 425.9 - 91.0 = 390.5 \, {S cm}^2 \, {mol}^{-1} \] Thus, the limiting molar conductance of acetic acid is \(390.5 \, {S cm}^2 \, {mol}^{-1}\).