Question

The limiting molar conductivity of La\(^{3+}\) and Cl\(^{-}\) ions in aqueous medium at 298 K are 209.10 $\times 10^{-4}$ and 76.35 $\times 10^{-4}$ S m\(^2\) mol\(^{-1}\), respectively. The transport number of Cl\(^{-}\) in an infinitely dilute aqueous solution of LaCl\(_3\) at 298 K is ................. (Round off to two decimal places)

Hint:

The transport number of an ion can be calculated by dividing its limiting molar conductivity by the total limiting molar conductivity of the electrolyte.

Answer 1

Correct Answer: 0.51

Step 1: Formula for transport number. 
The transport number (\(t\)) of an ion is given by the formula: \[ t_{\text{ion}} = \frac{\lambda_{\text{ion}}}{\lambda_{\text{total}}} \] where \(\lambda_{\text{ion}}\) is the limiting molar conductivity of the ion and \(\lambda_{\text{total}}\) is the total limiting molar conductivity of the electrolyte. 
Step 2: Limiting molar conductivity of LaCl\(_3\). 
For LaCl\(_3\), the total molar conductivity is the sum of the conductivities of La\(^{3+}\) and Cl\(^-\): \[ \lambda_{\text{total}} = \lambda_{\text{La}^{3+}} + \lambda_{\text{Cl}^-} = 209.10 \times 10^{-4} + 76.35 \times 10^{-4} = 285.45 \times 10^{-4} \] Step 3: Transport number of Cl\(^-\). 
Now, the transport number of Cl\(^-\) is: \[ t_{\text{Cl}^-} = \frac{76.35 \times 10^{-4}}{285.45 \times 10^{-4}} = 0.267 \] Rounding to two decimal places: \[ t_{\text{Cl}^-} = 0.27 \] Step 4: Conclusion. 
The transport number of Cl\(^-\) in an infinitely dilute aqueous solution of LaCl\(_3\) at 298 K is 0.27.