Answer the following: The molar conductivity at infinite dilution for and for respectively…

Formula: $\Lambda^\circ_m(\text{CH}_3\text{COOH}) = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{CH}_3\text{COO}^-}$ $\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$ Substitution: $\Lambda^\circ_m = 348\cdot65 + 41\cdot4 = 390\cdot05\text{ S cm}^2\text{mol}^{-1}$ $\alpha = \frac{40\cdot65\text{ S cm}^2\text{mol}^{-1}}{390\cdot05\text{ S cm}^2\text{mol}^{-1}} = 0\cdot1042\text{ (or } 10\cdot42\%\text{)}$ Final answer: 0.1042

Compound [B] is the weak electrolyte. Reason: A weak electrolyte is only partially dissociated at normal concentrations. Upon dilution, its degree of dissociation ($\alpha$) increases significantly according to Ostwald's dilution law, leading to a large increase in the number of conducting ions and hence a steep increase in molar conductivity (30 times). In contrast, a strong electrolyte is already completely dissociated, and dilution causes only a modest increase in molar conductivity (3 times) due to reduced interionic attractions.

No, copper sulphate solution cannot be stored in a zinc pot. Reason: $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0\cdot76\text{ V}$ is more negative than $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0\cdot34\text{ V}$, meaning zinc is more reactive and a stronger reducing agent than copper. Zinc spontaneously displaces copper from $\text{CuSO}_4$: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \longrightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$ Since $E^\circ_{\text{cell}} = +0\cdot34 - (-0\cdot76) = +1\cdot10\text{ V} > 0$, the reaction is thermodynamically spontaneous, and the zinc pot will dissolve.