JEE Main 2025 — Electrochemistry Question with Solution

Resistance and Resistivity:

$R=\rho \frac{\ell }{A}$

Where $R$ is resistance, $\rho$ is resistivity, $\ell$ is the length, and $A$ is the cross-sectional area of the cell. The term $\frac{\ell }{A}$ is known as the cell constant, $G^{\ast }$.

Conductivity and Conductance:

$\kappa =G\cdot G^{\ast }$

$G=\frac{1}{R};\kappa =\frac{1}{\rho }$

Where $G$ is conductance and $\kappa$ is conductivity.

Molar Conductivity:

${\lambda }_m=\frac{\kappa \times 1000}{C}$

Where ${\lambda }_m$ is molar conductivity and $C$ is concentration.

Given Information:

The resistance of the cell with the concentrated solution $R_c=100\Omega$.

We need to find the resistance $R_d$ for the dilute solution.

Steps to Solution:

The conductivity ratio for concentrated $({\kappa }_c)$ and dilute $({\kappa }_d)$ solutions relates to resistance:

$\frac{{\kappa }_c}{{\kappa }_d}=\frac{R_d}{R_c}$

From the plot or given values (assuming values for molar conductivity):

${\lambda }_{m,c}$ and ${\lambda }_{m,d}$ are the molar conductivities for the concentrated and dilute solutions.

If from the plot or data, ${\lambda }_{m,c}=150{\text{S cm}}^2{\text{mol}}^{-1}$ and ${\lambda }_{m,d}=100{\text{S cm}}^2{\text{mol}}^{-1}$.

Calculate the ratio of conductivities:

$\frac{{\kappa }_c}{{\kappa }_d}=\frac{\left({\lambda }_{m,c}\cdot C_c\right)}{\left({\lambda }_{m,d}\cdot C_d\right)}$

Substituting values into the equation:

$\frac{100\cdot (0.15{)}^2}{150\cdot (0.1{)}^2}=\frac{R_d}{100}$

Solving the equation gives:

$R_d=150\Omega$

Thus, the resistance of the conductivity cell with the dilute solution is $R_d=150\Omega$.