Question

Limiting molar conductivities for some ions in water at \(298\,K\) are given below: \[ \lambda_m^\circ(\mathrm{H^+})=349.6,\quad \lambda_m^\circ(\mathrm{Na^+})=50.1,\quad \lambda_m^\circ(\mathrm{Ca^{2+}})=119.0, \] \[ \lambda_m^\circ(\mathrm{Mg^{2+}})=106.0,\quad \lambda_m^\circ(\mathrm{Cl^-})=76.3,\quad \lambda_m^\circ(\mathrm{SO_4^{2-}})=160.0 \] Choose the correct decreasing order of molar conductivity \((\Lambda_m^\circ)\) for \(\mathrm{NaCl}\), \(\mathrm{HCl}\), \(\mathrm{CaCl_2}\) and \(\mathrm{MgSO_4}\).

• A. \(\mathrm{HCl} > \mathrm{CaCl_2} > \mathrm{MgSO_4} > \mathrm{NaCl}\)
• B. \(\mathrm{HCl} > \mathrm{CaCl_2} > \mathrm{NaCl} > \mathrm{MgSO_4}\)
• C. \(\mathrm{NaCl} > \mathrm{CaCl_2} > \mathrm{MgSO_4} > \mathrm{HCl}\)
• D. \(\mathrm{NaCl} > \mathrm{HCl} > \mathrm{CaCl_2} > \mathrm{MgSO_4}\)

Hint:

To find limiting molar conductivity of an electrolyte, simply add the ionic conductivities of all ions produced after dissociation, taking stoichiometric coefficients into account.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Molar conductivity depends on the ionic mobility of the ions in the solution. H\(^+\) ions have the highest ionic mobility due to the Grotthuss mechanism (proton hopping). Other ions contribute based on their charge and size.

Step 2: Detailed Explanation:
• HCl dissociates into H\(^+\) and Cl\(^-\). Due to the exceptionally high mobility of the H\(^+\) ion, HCl has the highest molar conductivity.
• CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2Cl\(^-\). The higher charge (2+) and the contribution of three ions overall result in higher conductivity than NaCl.
• NaCl dissociates into Na\(^+\) and Cl\(^-\). It has a standard ionic contribution.
• MgSO\(_4\) dissociates into Mg\(^{2+}\) and SO\(_4^{2-}\). While the ions are highly charged, their larger size and ionic association in solution often lead to lower molar conductivity than the simpler chlorides.

Step 3: Final Answer:
The order is HCl > CaCl\(_2\) > NaCl > MgSO\(_4\).