The molar conductivities \wedge _NaOAc^o and \wedge _HCl^o and at infinite dilution in water at 25oC are 91.0 and 426.2 Scm2/mol respectively. To calculate \wedge _HOAc^o , the additional value required is

Electrochemistry PYQ — JEE Main 2006 · Shift 0 · Q31 with Solution

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Correct answer: D

$\Lambda_m^\circ(AB)=\lambda_+^\circ+\lambda_-^\circ$

So,

$\Lambda_{\mathrm{NaOAc}}^\circ=\lambda_{\mathrm{Na}^+}^\circ+\lambda_{\mathrm{OAc}^-}^\circ=91.0$

$\Lambda_{\mathrm{HCl}}^\circ=\lambda_{\mathrm{H}^+}^\circ+\lambda_{\mathrm{Cl}^-}^\circ=426.2$

We need:

$\Lambda_{\mathrm{HOAc}}^\circ=\lambda_{\mathrm{H}^+}^\circ+\lambda_{\mathrm{OAc}^-}^\circ$

=\lambda_{\mathrm{Na}^+}^\circ+\lambda_{\mathrm{OAc}^-}^\circ+\lambda_{\mathrm{H}^+}^\circ+\lambda_{\mathrm{Cl}^-}^\circ$$ To get $\lambda_{\mathrm{H}^+}^\circ+\lambda_{\mathrm{OAc}^-}^\circ$, we must subtract $$\lambda_{\mathrm{Na}^+}^\circ+\lambda_{\mathrm{Cl}^-}^\circ$$ But this is exactly: $$\Lambda_{\mathrm{NaCl}}^\circ=\lambda_{\mathrm{Na}^+}^\circ+\lambda_{\mathrm{Cl}^-}^\circ$$ Therefore, $$\Lambda_{\mathrm{HOAc}}^\circ=\Lambda_{\mathrm{HCl}}^\circ+\Lambda_{\mathrm{NaOAc}}^\circ-\Lambda_{\mathrm{NaCl}}^\circ$$ 3. Hence, the additional value required is $$\boxed{\Lambda_{\mathrm{NaCl}}^\circ}$$ 4. Option check: - **A: $\Lambda_{H_2O}^\circ$** — not useful - **B: $\Lambda_{KCl}^\circ$** — contains $K^+$, not $Na^+$ - **C: $\Lambda_{NaOH}^\circ$** — contains $OH^-$, not $Cl^-$ - **D: $\Lambda_{NaCl}^\circ$** — correct So the correct option is **D**.