The equivalent conductivity (in Ω^-1 cm^2 eq^-1) of 1.0 M - solution of specific conductance 2.6 × 1 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Definition and Formula for Equivalent Conductivity Equivalent conductivity ($\Lambda_{eq}$) of an electrolyte solution is defined as the conducting power of all the ions produced by dissolving one gram-equivalent of the electrolyte in a given volume of solution. The relationship between equivalent conductivity ($\Lambda_{eq}$), specific conductance ($\kappa$), and normality ($N$) of the solution is expressed as: $$\Lambda_{eq} = \frac{\kappa \times 1000}{N}$$ Where: * $\Lambda_{eq}$ is the equivalent conductivity in $\text{ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$. * $\kappa$ is the specific conductance (conductivity) in $\text{ohm}^{-1}\text{ cm}^{-1}$. * $N$ is the normality of the solution in $\text{eq L}^{-1}$ (or $\text{N}$). * The factor $1000$ is a volume conversion factor with units $\text{cm}^3\text{ L}^{-1}$ used to convert liters into cubic centimeters. Step 2 - Determine the Normality of the \ce{H2SO4} Solution We are given a $1.0\text{ M}$ solution of sulphuric acid ($\ce{H2SO4}$). Normality ($N$) is related to molarity ($M$) by the valency factor ($n$-factor), which represents the number of replaceable hydrogen ions per molecule of the acid: $$N = M \times n\text{-factor}$$ Sulphuric acid ($\ce{H2SO4}$) is a strong, dibasic acid that releases two moles of hydrogen ions per mole of acid upon dissociation in water: $$\ce{H2SO4(aq) -> 2H^+(aq) + SO4^{2-}(aq)}$$ Therefore, the $n$-factor of $\ce{H2SO4}$ is $2$. Now, substitute the molarity into the equation to find the normality: $$N = 1.0\text{ M} \times 2 = 2.0\text{ N}$$ Step 3 - Calculate the Equivalent Conductivity ($\Lambda_{eq}$) We are given: * Specific conductance ($\kappa$) = $2.6 \times 10^{-1}\text{ ohm}^{-1}\text{ cm}^{-1} = 0.26\text{ ohm}^{-1}\text{ cm}^{-1}$ * Normality ($N$) = $2.0\text{ N}$ (calculated in Step 2) Substituting these values with their respective units into the equivalent conductivity equation: $$\Lambda_{eq} = \frac{\kappa \times 1000}{N}$$ $$\Lambda_{eq} = \frac{\left(2.6 \times 10^{-1}\text{ ohm}^{-1}\text{ cm}^{-1}\right) \times 1000\text{ cm}^3\text{ L}^{-1}}{2.0\text{ eq L}^{-1}}$$ $$\Lambda_{eq} = \frac{260\text{ ohm}^{-1}\text{ cm}^{-1}\text{ cm}^3\text{ L}^{-1}}{2.0\text{ eq L}^{-1}}$$ $$\Lambda_{eq} = 130\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$$ Expressing this result in standard scientific notation yields: $$\Lambda_{eq} = \mathbf{1.3 \times 10^2\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}}$$ Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** As mathematically calculated, the equivalent conductivity of the $1.0\text{ M}$ sulphuric acid solution is exactly $1.3 \times 10^2\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$ (or $130\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$). * **Option (B) is incorrect:** This value ($6.5 \times 10^1 = 65\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$) is obtained if the basicity of $\ce{H2SO4}$ is incorrectly taken as $4$ or if the correct result is halved due to a computational error. * **Option (C) is incorrect:** This value ($1.3 \times 10^{-1} = 0.13\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$) represents an error in the decimal place where division by $1000$ was done instead of multiplication. * **Option (D) is incorrect:** This value ($2.6 \times 10^2 = 260\text{ ohm}^{-1}\text{ cm}^2\text{ eq}^{-1}$) represents the molar conductivity ($\Lambda_m = \frac{\kappa \times 1000}{M}$) of the solution. This is obtained if one forgets to convert molarity into normality and instead uses the molarity of $1.0\text{ M}$ as the denominator. $$\text{Correct Option: } \boxed{\text{A}}$$