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)
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is:
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: