Question

Equivalent conductance of $ NaCl, HCl$ and $C_2H_5COONa $ at infinete dilution are $126.45, 426.16$ and $91 \Omega ^{-1} cm^2 $ , respectively. The equivalent conductance of $C_2H_5COOH $ is

• A. 201.28 $\Omega^{-1} \, cm^2 $
• B. 390.71 $\Omega^{-1} \, cm^2 $
• C. 698.28 $\Omega^{-1} \, cm^2 $
• D. 540.48 $\Omega^{-1} \, cm^2 $

Answer 1

The Correct Option is B

To find the equivalent conductance of C_2H_5COOH, we will use the law of independent migration of ions. According to this law, the equivalent conductance of an electrolyte at infinite dilution is the sum of the equivalent conductances of its constituent ions.

We are given:

• Equivalent conductance of NaCl (\Lambda^\infty_{NaCl}) = 126.45 \, \Omega^{-1} \, cm^2
• Equivalent conductance of HCl (\Lambda^\infty_{HCl}) = 426.16 \, \Omega^{-1} \, cm^2
• Equivalent conductance of C_2H_5COONa (\Lambda^\infty_{C_2H_5COONa}) = 91 \, \Omega^{-1} \, cm^2

The reaction involved when mixing these substances is:

• C_2H_5COONa + HCl \rightarrow C_2H_5COOH + NaCl

Using the law of independent migration of ions, we can calculate the equivalent conductance of C_2H_5COOH at infinite dilution as follows:

\Lambda^\infty_{C_2H_5COOH} = \Lambda^\infty_{HCl} + \Lambda^\infty_{C_2H_5COONa} - \Lambda^\infty_{NaCl}

Substituting the given values:

\Lambda^\infty_{C_2H_5COOH} = 426.16 + 91 - 126.45

Calculating this gives:

\Lambda^\infty_{C_2H_5COOH} = 390.71 \, \Omega^{-1} \, cm^2

Therefore, the equivalent conductance of C_2H_5COOH at infinite dilution is 390.71 \Omega^{-1} \, cm^2, which matches option B.