If the values of Lambda_{infty} of NH_4Cl, Na | vedclass

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(A) According to Kohlrausch's law of independent migration of ions:$\Lambda_{\infty} (NH_4Cl) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (Cl^-) =

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$238$

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(A) According to Kohlrausch's law of independent migration of ions:$\Lambda_{\infty} (NH_4Cl) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (Cl^-) = 130 \ \Omega^{-1} \ cm^2 \ equiv^{-1} \ (i)$$\Lambda_{\infty} (NaOH) = \Lambda_{\infty} (Na^+) + \Lambda_{\infty} (OH^-) = 217 \ \Omega^{-1} \ cm^2 \ equiv^{-1} \ (ii)$$\Lambda_{\infty} (NaCl) = \Lambda_{\infty} (Na^+) + \Lambda_{\infty} (Cl^-) = 109 \ \Omega^{-1} \ cm^2 \ equiv^{-1} \ (iii)$To find $\Lambda_{\infty} (NH_4OH)$,we perform the operation: $(i) + (ii) - (iii)$$\Lambda_{\infty} (NH_4OH) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (Cl^-) + \Lambda_{\infty} (Na^+) + \Lambda_{\infty} (OH^-) - (\Lambda_{\infty} (Na^+) + \Lambda_{\infty} (Cl^-))$$\Lambda_{\infty} (NH_4OH) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (OH^-)$$\Lambda_{\infty} (NH_4OH) = 130 + 217 - 109 = 238 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$

If the values of $\Lambda_{\infty}$ of $NH_4Cl, NaOH$ and $NaCl$ are $130, 217$ and $109 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$ respectively,the $\Lambda_{\infty}$ of $NH_4OH$ in $\Omega^{-1} \ cm^2 \ equiv^{-1}$ is

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