Q29 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
Given below are 4 equations showing Molar conductivities at infinite dilution of various electrolytes. Which one of them represents the correct equation?
$\begin{aligned}
& \left(\Lambda_m^0\right)_{N a B r}-\left(\Lambda_m^0\right)_{N a C l}=\left(\Lambda_m^0\right)_{\mathrm{KCl}}-\left(\Lambda_m^0\right)_{\mathrm{KBr}} \\
& \left(\Lambda_m^0\right)_{\mathrm{HCl}}+\left(\Lambda_m^0\right)_{\mathrm{KOH}}-\left(\Lambda_m^0\right)_{\mathrm{KCl}}=\left(\Lambda_m^0\right)_{\mathrm{H}_2 \mathrm{O}} \\
& \left(\Lambda_m^0\right)_{\mathrm{KBr}}-\left(\Lambda_m^0\right)_{\mathrm{NaBr}}=\left(\Lambda_m^0\right)_{\mathrm{NaBr}}-\left(\Lambda_m^0\right)_{\mathrm{Nal}} \\
& \left(\Lambda_m^0\right)_{N H_4 \mathrm{Cl}}-\left(\Lambda_m^0\right)_{\mathrm{NH}_4 \mathrm{NO}_3}=\left(\Lambda_m^0\right)_{N H_4 \mathrm{Cl}}-\left(\Lambda_m^0\right)_{N H 4 B r}
\end{aligned}$
- AEquation 4
- BEquation 1
- Correct answerCEquation 2
- DEquation 3
Solution
Using Kohlrausch's law, expanding Equation 2 LHS: $(A^0_m)_{HCl} + (A^0_m)_{KOH} - (A^0_m)_{KCl}$
$= [\lambda^0(H^+) + \lambda^0(Cl^-)] + [\lambda^0(K^+) + \lambda^0(OH^-)] - [\lambda^0(K^+) + \lambda^0(Cl^-)]$
$= \lambda^0(H^+) + \lambda^0(OH^-)$
$= (A^0_m)_{H_2O}$ = RHS
Equation 2 is correct. The cancellation of $\lambda^0(Cl^-)$ and $\lambda^0(K^+)$ on both sides confirms LHS = RHS.
$= [\lambda^0(H^+) + \lambda^0(Cl^-)] + [\lambda^0(K^+) + \lambda^0(OH^-)] - [\lambda^0(K^+) + \lambda^0(Cl^-)]$
$= \lambda^0(H^+) + \lambda^0(OH^-)$
$= (A^0_m)_{H_2O}$ = RHS
Equation 2 is correct. The cancellation of $\lambda^0(Cl^-)$ and $\lambda^0(K^+)$ on both sides confirms LHS = RHS.
Hence, the correct answer is option: 'C'.