The resistance of 1 M - CH3COOH solution is 250 Ω, when measured in a cell of cell constant 125 m^-1 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Given Parameters and Understand the Units We are given the following experimental values for an aqueous solution of acetic acid ($\ce{CH3COOH}$): * Molar concentration of the solution ($C$) = $1\text{ M} = 1\text{ mol/L}$ * Measured electrical resistance ($R$) = $250\ \Omega$ * Cell constant of the conductivity cell ($G^* = \frac{l}{A}$) = $125\text{ m}^{-1}$ Our goal is to calculate the molar conductivity ($\Lambda_m$) in SI units, specifically in $\Omega^{-1}\text{ m}^2\text{ mol}^{-1}$. Step 2 - Convert the Concentration ($C$) into SI Units ($\text{mol/m}^3$) The given concentration is in molarity ($\text{mol/L}$). To perform calculations strictly in SI units, we must convert liters ($\text{L}$) to cubic meters ($\text{m}^3$). Using the standard volumetric conversion: $$1\text{ m}^3 = 1000\text{ L} \implies 1\text{ L} = 10^{-3}\text{ m}^3$$ Now, substitute this conversion factor to find the concentration in $\text{mol/m}^3$: $$C = 1\text{ mol/L} = \frac{1\text{ mol}}{10^{-3}\text{ m}^3}$$ $$C = 1000\text{ mol/m}^3$$ Step 3 - Calculate the Specific Conductivity (Conductivity, $\kappa$) The specific conductivity ($\kappa$) represents the conducting capability of a unit volume of the electrolyte. It is related to the electrical resistance ($R$) and the cell constant ($G^*$) by the formula: $$\kappa = \frac{G^*}{R}$$ Substitute the given values with their units into the equation: $$\kappa = \frac{125\text{ m}^{-1}}{250\ \Omega}$$ $$\kappa = 0.5\ \Omega^{-1}\text{ m}^{-1}$$ Step 4 - Calculate the Molar Conductivity ($\Lambda_m$) Molar conductivity ($\Lambda_m$) is defined as the conductance of a volume of solution containing exactly one mole of the electrolyte. In SI units, the formula is: $$\Lambda_m = \frac{\kappa}{C}$$ Substitute the calculated specific conductivity ($\kappa = 0.5\ \Omega^{-1}\text{ m}^{-1}$) and the concentration in SI units ($C = 1000\text{ mol/m}^3$): $$\Lambda_m = \frac{0.5\ \Omega^{-1}\text{ m}^{-1}}{1000\text{ mol/m}^3}$$ $$\Lambda_m = \boxed{5.0 \times 10^{-4}\ \Omega^{-1}\text{ m}^2\text{ mol}^{-1}}$$ Step 5 - Evaluate the Options * **Option (A) is correct:** As calculated, the molar conductivity is exactly $5.0 \times 10^{-4}\ \Omega^{-1}\text{ m}^2\text{ mol}^{-1}$. * **Option (B) is incorrect:** This value ($500$) is obtained if the units of the cell constant are ignored or misconverted during the calculation. * **Option (C) is incorrect:** This value ($2 \times 10^{-3}$) is mathematically incorrect and represents a division error where resistance and cell constant are inverted. * **Option (D) is incorrect:** This value ($200$) is far from the calculated SI value and results from a dimensional error. $$\text{Correct Option: } \boxed{\text{A}}$$