Question

The specific conductivity of a solution containing 1.0 g of anhydrous BaCl\(_2\) in 200 cm\(^3\) of the solution has been found to be 0.0058 Scm\(^{-1}\). The molar and equivalent conductivity of the solution respectively are:

• A. 120.83 \, \(cm^2\) \, \(eq^{-1}\) \, \text{and} \, 241.67 \, \(cm^2\) \, \(eq^{-1}\)
• B. 150.5 \, \(cm^2\) \, \(eq^{-1}\) \, \text{and} \, 289.7 \, \(cm^2\) \, \(eq^{-1}\)
• C. 241.6 \, \(cm^2\) \, \(mol^{-1}\) \, \text{and} \, 120.83 \, \(cm^2\) \, \(eq^{-1}\)
• D. 248.6 \, \(cm^2\) \, \(mol^{-1}\) \, \text{and} \, 180.3 \, \(cm^2\) \, \(eq^{-1}\)

Hint:

Use the conductivity equation to calculate the molar and equivalent conductivities of solutions, especially when given the specific conductivity and concentration.

Answer 1

The Correct Option is C

The molar conductivity is given by: \[ \Lambda_m = \frac{\kappa}{C} \] where \( \kappa \) is the conductivity and \( C \) is the concentration. For BaCl\(_2\), the equivalent conductivity \( \Lambda_e \) is calculated as: \[ \Lambda_e = \frac{\kappa}{C_{\text{eq}}} \] By substituting the given values, we can calculate the molar and equivalent conductivities. The correct values match Option (C).