Question
At 298, the limited molar conductivity at infinite dilution of sodium benzoate, hydrochloric acid and sodium chloride are 8.24 ´ 10-3 S m2 mol-1, 4.26 ´ 10-2 S m2 mol-1 and 1.26 ´ 10-2 respectively. Calculate the limiting molar conductivity of aqueous benzoic acid.
Answer :
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Skip in 10
We are asked to calculate the limiting molar conductivity of aqueous benzoic acid ($\Lambda_m^0 (\text{C}_6\text{H}_5\text{COOH})$) using the given limiting molar conductivities of ions. Let's solve this step by step manually. --- ### Given data: | Substance | $\Lambda_m^0$ (S·m²·mol⁻¹) | | ------------------------------- | -------------------------- | | Sodium benzoate ($C_6H_5COONa$) | $8.24 \times 10^{-3}$ | | Hydrochloric acid ($HCl$) | $4.26 \times 10^{-2}$ | | Sodium chloride ($NaCl$) | $1.26 \times 10^{-2}$ | We are asked to find $\Lambda_m^0 (\text{C}_6\text{H}_5\text{COOH})$. --- ### Step 1: Recall the relation For an ionic compound, the limiting molar conductivity is the sum of the contributions of its ions: $$ \Lambda_m^0 (\text{salt}) = \lambda^0_{\text{cation}} + \lambda^0_{\text{anion}} $$
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