Equivalence conductance at infinite dilution of NH4Cl, NaOH and NaCl are 129.8, 217.4 and , respecti — Electrochemistry Chemistry Question
Question
Equivalence conductance at infinite dilution of NH4Cl, NaOH and NaCl are 129.8, 217.4 and $108.9 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$, respectively. If the equivalent conductance of 0.01 N solution of NH4OH is $9.532 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$, then the degree of dissociation of NH4OH at this temperature is
Answer: C
💡 Solution & Explanation
By Kohlrausch's Law, $\Lambda^\circ_{eq}(NH_4OH) = \Lambda^\circ_{eq}(NH_4Cl) + \Lambda^\circ_{eq}(NaOH) - \Lambda^\circ_{eq}(NaCl) = 129.8 + 217.4 - 108.9 = 238.3 \text{ }\Omega^{-1}\text{cm}^2\text{eq}^{-1}$. Degree of dissociation $\alpha = \Lambda_{eq}^c / \Lambda^\circ_{eq} = 9.532 / 238.3 = 0.04$. Percentage dissociation = $0.04 \times 100 = 4.0\%$. Therefore, correct answer is C.
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