Question

The resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.3 S m⁻¹. If resistance of 0.4 M solution of the same electrolyte is 260 Ω, its molar conductivity is:

[HINT: $\kappa$ = (1/R) G*, \Lambda$$_{m} = ($\kappa$ × 1000) / M, take care of units]

• A. 6.25 × 10⁻⁴ S m² mol⁻¹
• B. 6.25 × 10⁻² S m² mol⁻¹
• C. 62.5 S m² mol⁻¹
• D. 625 × 10⁻⁴ S m² mol⁻¹

Answer 1

Answer: (A)

6.25 × 10⁻⁴ S m² mol⁻¹

Answer 2

Here's how to solve this problem:

1. *Calculate the cell constant (G)**:

   Using the data for the 0.2 M solution:

   $\kappa_1 = (1/R_1) G^*$

   $1.3 \, S \, m^{-1} = (1/50 \, \Omega) G^*$

   $G^* = 1.3 \, S \, m^{-1} \times 50 \, \Omega = 65 \, m^{-1}$

2. Calculate the specific conductance ($\kappa_2$) of the 0.4 M solution:

   $\kappa_2 = (1/R_2) G^*$

   $\kappa_2 = (1/260 \, \Omega) \times 65 \, m^{-1}$

   $\kappa_2 = 0.25 \, S \, m^{-1}$

3. Calculate the molar conductivity ($\Lambda_{m2}$) of the 0.4 M solution:

   $\Lambda_{m} = \kappa / C$, where C is the molar concentration in $mol \, m^{-3}$.

   Convert molarity from $mol \, L^{-1}$ to $mol \, m^{-3}$:

   $C_2 = 0.4 \, mol \, L^{-1} \times 1000 \, L \, m^{-3} = 400 \, mol \, m^{-3}$

   $\Lambda_{m2} = \kappa_2 / C_2$

   $\Lambda_{m2} = 0.25 \, S \, m^{-1} / 400 \, mol \, m^{-3}$

   $\Lambda_{m2} = 6.25 \times 10^{-4} \, S \, m^2 \, mol^{-1}$

Therefore, the molar conductivity of the 0.4 M solution is $6.25 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.