The relationship between molar conductivity $\Lambda_m$, molar conductivity at infinite dilution $\Lambda_m^\circ$, and concentration $C$ for a weak electrolyte can be derived from the dissociation equilibrium. The correct equation involves the dissociation constant $K_a$ and accounts for the variation of $\Lambda_m$ with concentration. For weak electrolytes, the molar conductivity $\Lambda_m$ is related to the degree of dissociation $\alpha$ as:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ}. \]
The dissociation constant $K_a$ is expressed as:
\[ K_a = \frac{C\alpha^2}{1 - \alpha}. \]
Substituting $\alpha =\frac{\Lambda_m}{\Lambda_m^\circ}$ into the equation:
\[ K_a = \frac{C \left(\frac{\Lambda_m}{\Lambda_m^\circ}\right)^2}{1 - \frac{\Lambda_m}{\Lambda_m^\circ}}. \]
Simplifying and rearranging, the equation becomes:
\[ \Lambda_m^2 C - K_a \Lambda_m^{\circ 2} + K_a \Lambda_m \Lambda_m^\circ = 0. \]
This is the equation that correctly represents the relationship between molar conductivity, concentration, and dissociation constant for a weak electrolyte.
Final Answer: (1)
If the molar conductivity ($\Lambda_m$) of a 0.050 mol $L^{–1}$ solution of a monobasic weak acid is 90 S $cm^{2} mol^{–1}$, its extent (degree) of dissociation will be:
[Assume: $\Lambda^0$ = 349.6 S $cm^{2} mol^{–1}$ and $\Lambda^0_{\text{acid}}$ = 50.4 S$ cm^{2} mol^{–1}$]
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.