Understanding Cell Constant and Molar Conductance

The question involves concepts of resistance, specific conductance, and molar conductance of electrolyte solutions, measured using an electrolytic cell. A key property of the cell used for measurements is its cell constant ($G^*$). The cell constant depends only on the geometry of the cell (distance between electrodes and area of electrodes) and remains constant for a given cell.

The relationship between resistance ($R$), specific conductance ($\kappa$), and cell constant ($G^*$) is given by:

$\kappa = \frac{1}{R} \times G^*$

Or, $G^* = R \times \kappa$

Molar conductance ($\Lambda_m$) relates specific conductance ($\kappa$) to the molar concentration ($C$) of the electrolyte. The formula for molar conductance is:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

where $\kappa$ is in S cm$^{-1}$, $C$ is in mol L$^{-1}$ (or M), and $\Lambda_m$ is in S cm$^2$ mol$^{-1}$. The factor of 1000 converts the volume from L to cm$^3$ (since 1 L = 1000 cm$^3$) to match the units of $\kappa$.

Cell Constant Calculation using KCl Solution

We are given the resistance and specific conductance of a 0.1 M KCl solution in the cell:

• Resistance of KCl solution, $R_{KCl} = 300$ ohms
• Specific conductance of KCl solution, $\kappa_{KCl} = 1.5$ S cm$^{-1}$

Using the formula $G^* = R \times \kappa$, we can calculate the cell constant:

$G^* = R_{KCl} \times \kappa_{KCl}$

$G^* = 300 \, \Omega \times 1.5 \, \text{S cm}^{-1}$

$G^* = 450 \, \text{cm}^{-1}$

Since the same cell is used for the NaCl solution, the cell constant for the NaCl measurement is also $450 \, \text{cm}^{-1}$.

Specific Conductance of NaCl Solution

We are given the resistance of the 0.05 M NaCl solution in the same cell:

• Resistance of NaCl solution, $R_{NaCl} = 750$ ohms

Using the relationship $\kappa = \frac{1}{R} \times G^*$, we can find the specific conductance of the NaCl solution:

$\kappa_{NaCl} = \frac{G^*}{R_{NaCl}}$

$\kappa_{NaCl} = \frac{450 \, \text{cm}^{-1}}{750 \, \Omega}$

$\kappa_{NaCl} = 0.6 \, \text{S cm}^{-1}$

Molar Conductance of NaCl Solution

Now we have the specific conductance of the NaCl solution and its molar concentration:

• Specific conductance of NaCl solution, $\kappa_{NaCl} = 0.6$ S cm$^{-1}$
• Molar concentration of NaCl solution, $C_{NaCl} = 0.05$ M (or mol L$^{-1}$)

Using the formula for molar conductance $\Lambda_m = \frac{\kappa \times 1000}{C}$:

$\Lambda_{m, NaCl} = \frac{\kappa_{NaCl} \times 1000}{C_{NaCl}}$

$\Lambda_{m, NaCl} = \frac{0.6 \, \text{S cm}^{-1} \times 1000 \, \text{cm}^3/\text{L}}{0.05 \, \text{mol/L}}$

$\Lambda_{m, NaCl} = \frac{600}{0.05} \, \text{S cm}^2 \text{mol}^{-1}$

$\Lambda_{m, NaCl} = 12000 \, \text{S cm}^2 \text{mol}^{-1}$

Thus, the molar conductance of the 0.05 M NaCl solution is 12,000 S cm$^2$ mol$^{-1}$.

Summary of Steps

1. Calculate the cell constant ($G^*$) using the given data for the KCl solution ($G^* = R_{KCl} \times \kappa_{KCl}$).
2. Calculate the specific conductance ($\kappa$) of the NaCl solution using the calculated cell constant and the given resistance of the NaCl solution ($\kappa_{NaCl} = G^* / R_{NaCl}$).
3. Calculate the molar conductance ($\Lambda_m$) of the NaCl solution using its specific conductance and molar concentration ($\Lambda_{m, NaCl} = \frac{\kappa_{NaCl} \times 1000}{C_{NaCl}}$).