Question

What is the molar conductivity at infinite dilution of \(\text{CaCl}_2\), if the molar conductivity of \(\text{Ca}^{2+}\) ion and \(\text{Cl}^-\) ion at infinite dilution is \(119\) and \(71~\Omega^{-1}\text{cm}^2\text{mol}^{-1}\) respectively?

• A. \(431.0~\Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
• B. \(341.0~\Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
• C. \(261.0~\Omega^{-1}\text{cm}^2\text{mol}^{-1}\)
• D. \(126.0~\Omega^{-1}\text{cm}^2\text{mol}^{-1}\)

Hint:

Always multiply the ionic conductivity by the number of ions produced during dissociation.

Answer 1

The Correct Option is C

Step 1: Write the formula for molar conductivity at infinite dilution.
According to Kohlrausch’s law: \[ \Lambda_m^0 = \sum \nu_i \lambda_i^0 \] Step 2: Apply the formula to \(\text{CaCl}_2\).
\(\text{CaCl}_2\) dissociates as: \[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \] Step 3: Substitute the given values.
\[ \Lambda_m^0 = 1(119) + 2(71) \] \[ \Lambda_m^0 = 119 + 142 = 261~\Omega^{-1}\text{cm}^2\text{mol}^{-1} \] Step 4: Final conclusion.
Hence, the molar conductivity of \(\text{CaCl}_2\) at infinite dilution is \(261.0~\Omega^{-1}\text{cm}^2\text{mol}^{-1}\).