Question

Resistance of a conductivity cell filled with \(0.1 \text{ mol L}^{-1}\) NaCl is \(100 \, \Omega\). If the resistance of the same cell when filled with \(0.02 \text{ mol L}^{-1}\) NaCl solution is \(258 \, \Omega\), then the conductivity of \(0.02 \text{ mol L}^{-1}\) NaCl solution is:
\[ \text{(Conductivity of }0.1 \text{ mol L}^{-1}\text{ NaCl} = 1.29 \, \text{S m}^{-1}) \]

• A. \(1.0 \, \text{S m}^{-1}\)
• B. \(0.2 \, \text{S m}^{-1}\)
• C. \(2.0 \, \text{S m}^{-1}\)
• D. \(0.5 \, \text{S m}^{-1}\)

Hint:

For the same conductivity cell, cell constant remains unchanged, so: \[ \kappa R = \text{constant} \] Conductivity is inversely proportional to resistance.

Answer 1

The Correct Option is D

Step 1: Recall the cell constant formula.
For a conductivity cell: Cell constant $G^* = \kappa \times R$, where $\kappa$ = conductivity (S/m or S/cm) and $R$ = resistance ($\Omega$). The cell constant is fixed for a given cell regardless of the solution.
Step 2: Use the known conductivity of 0.1 mol/L NaCl.
The standard conductivity of 0.1 mol/L NaCl is $\kappa_1 = 1.29$ S/m (a standard reference value). The resistance measured = 100 $\Omega$. Cell constant: \[ G^* = \kappa_1 \times R_1 = 1.29 \times 100 = 129 \text{ m}^{-1} \]
Step 3: Use the cell constant to find $\kappa_2$ for 0.02 mol/L NaCl.
The same cell with 0.02 mol/L NaCl gives resistance $R_2 = 258$ $\Omega$. Since $G^*$ is constant: \[ \kappa_2 = \frac{G^*}{R_2} = \frac{129}{258} = 0.5 \text{ S/m} \]
Step 4: Verify the ratio makes physical sense.
The concentration decreased from 0.1 mol/L to 0.02 mol/L (a factor of 5 decrease). Conductivity should decrease proportionally (at low concentrations): $1.29/5 \approx 0.258$ S/m. But resistance increased by factor 258/100 = 2.58, giving $1.29/2.58 = 0.5$ S/m. The molar conductivity actually increases at lower concentration (Kohlrausch's law), so 0.5 S/m is reasonable.
Step 5: Cross-check units.
$G^*$ in m$^{-1}$, $R$ in $\Omega$, $\kappa = G^*/R$ in m$^{-1}$/$\Omega$ = S/m. Consistent.
Step 6: State the final answer.
Conductivity of 0.02 mol/L NaCl in this cell = $\frac{129}{258} = 0.5$ S/m.
\[ \boxed{\kappa = 0.5 \text{ S/m (option 4)}} \]