Question
For a salt $\mathbf{XY}$, which is a strong electrolyte, the plot of $\Lambda_{m}$ versus $\sqrt{c}$ has a slope of $- 90.0\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 3/2}$ $L^{1/2}$ at 298 K. At 0.01 M concentration of XY, the value of $\Lambda_{m}$ is $145.0\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}$. The limiting molar conductivity of $\mathbf{Y}^{-}$ion $\left( \lambda_{\mathbf{Y}^{-}}^{0} \right.\ $, in $\left. \ S{cm}^{2}{\text{ }mol}^{- 1} \right)$ at 298 K will be
(Given : $\lambda_{X^{+}}^{0} = 74.0\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}$ )
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Solution
Sol. [Key Concept: Λm vs √c for a strong electrolyte (Debye–Hückel–Onsager)]
$y = mx + c$
slope = -90
According to $\lambda_{m} = \lambda_{m}^{0} - A\sqrt{c}$
$$\begin{matrix} & 145 = \lambda_{m}^{0} - 90 \times 0.1 \\ & \lambda_{m}^{\circ} = 154\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1} \\ & \lambda_{m(XY)}^{\circ} = \lambda_{X^{+}}^{\circ} + \lambda_{Y^{-}}^{\circ} \\ & 154 = 74 + \lambda_{Y^{-}}^{\circ} \\ & \lambda_{Y^{-}}^{\circ} = 80\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1} \end{matrix}$$
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