The resistance of a solution A is 50 Ω and that of solution B is 100 Ω, both solutions being taken i — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Relation Between Conductivity and Resistance The specific conductance or conductivity ($\kappa$) of an electrolyte solution is related to its measured resistance ($R$) and the cell constant ($G^*$) of the conductivity cell by the formula: $$\kappa = \frac{G^*}{R}$$ Where: * $\kappa$ is the specific conductance of the solution in $\text{ohm}^{-1}\text{ cm}^{-1}$ (or $\text{S cm}^{-1}$). * $G^*$ is the cell constant of the conductivity cell in $\text{cm}^{-1}$. * $R$ is the electrical resistance of the solution in $\text{ohm}$ ($\Omega$). Step 2 - Express the Conductivity of Both Solutions Using the cell constant $G^*$ for the same conductivity cell, we can write the individual conductivities of solution A and solution B: * **For Solution A ($R_A = 50\ \Omega$):** $$\kappa_A = \frac{G^*}{50\ \Omega}$$ * **For Solution B ($R_B = 100\ \Omega$):** $$\kappa_B = \frac{G^*}{100\ \Omega}$$ Step 3 - Determine the Dilution Effect Upon Mixing Equal Volumes When equal volumes ($V$) of solution A and solution B are mixed, the total volume of the resulting mixture becomes $2V$. Assuming there is no change in the degree of dissociation of either solution on mixing (as stated in the question): * The concentration of ions from solution A is halved due to doubling of the volume. * The concentration of ions from solution B is also halved. Since specific conductance is directly proportional to the concentration of charge-carrying ions in the solution, the contribution of each component to the total conductivity of the mixture is halved. Therefore, the specific conductance of the mixture ($\kappa_{\text{mix}}$) is the sum of these diluted conductivities: $$\kappa_{\text{mix}} = \frac{\kappa_A}{2} + \frac{\kappa_B}{2} = \frac{\kappa_A + \kappa_B}{2}$$ Step 4 - Calculate the Conductivity and Resistance of the Mixture Substitute the expressions for $\kappa_A$ and $\kappa_B$ into our formula for $\kappa_{\text{mix}}$: $$\kappa_{\text{mix}} = \frac{\frac{G^*}{50} + \frac{G^*}{100}}{2}$$ $$\kappa_{\text{mix}} = \frac{1}{2} \cdot G^* \left(\frac{2 + 1}{100}\right)$$ $$\kappa_{\text{mix}} = \frac{3}{200} G^*$$ The resistance of the mixture ($R_{\text{mix}}$) in the same conductivity cell (having the same cell constant $G^*$) is given by: $$R_{\text{mix}} = \frac{G^*}{\kappa_{\text{mix}}}$$ Substitute the calculated value of $\kappa_{\text{mix}}$: $$R_{\text{mix}} = \frac{G^*}{\frac{3}{200} G^*}$$ $$R_{\text{mix}} = \frac{200}{3}\ \Omega$$ $$R_{\text{mix}} = \mathbf{66.67\ \Omega}$$ Step 5 - Evaluate the Options * **Option (A) is incorrect:** $150\ \Omega$ is the sum of the two resistances ($R_A + R_B$). This would represent two cells connected in series, rather than mixing the solutions. * **Option (B) is incorrect:** $75\ \Omega$ is the simple arithmetic mean of the two resistances ($\frac{50 + 100}{2}$). However, resistances of parallel/mixed conductors do not average arithmetically. * **Option (C) is incorrect:** $33.33\ \Omega$ is the equivalent resistance if the two solutions were mixed without any dilution effect (i.e., if we managed to put them in parallel while keeping their concentrations constant, which is physically impossible when mixing equal volumes as the total volume doubles). * **Option (D) is correct:** As shown by our derivation, taking the dilution effect into account yields a mixture resistance of exactly $66.67\ \Omega$. $$\text{Correct Option: } \boxed{\text{D}}$$