What is the percent dissociation of NH_4OH if | vedclass

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(D) According to Kohlrausch's law,the molar conductivity at infinite dilution for $NH_4OH$ is given by: $\Lambda^0_{m}(NH_4OH) = \Lambda^0_{m}(NH_4Cl)

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$\frac{100}{26}$

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(D) According to Kohlrausch's law,the molar conductivity at infinite dilution for $NH_4OH$ is given by: $\Lambda^0_{m}(NH_4OH) = \Lambda^0_{m}(NH_4Cl) + \Lambda^0_{m}(NaOH) - \Lambda^0_{m}(NaCl)$.Substituting the given values: $\Lambda^0_{m}(NH_4OH) = 130 + 213 - 109 = 234 \ S \ cm^2 \ mol^{-1}$.The degree of dissociation $(\alpha)$ is calculated as: $\alpha = \frac{\Lambda^m_c}{\Lambda^0_m} = \frac{9.0}{234}$.The percent dissociation is: $\alpha 	imes 100 = \frac{9.0 	imes 100}{234} = \frac{900}{234} = \frac{100}{26} \%$.

What is the percent dissociation of $NH_4OH$ if the molar conductance at zero concentration for $NH_4Cl, NaCl$,and $NaOH$ are $130, 109$,and $213 \ S \ cm^2 \ mol^{-1}$ respectively,and the molar conductivity of $0.01 \ M \ NH_4OH$ is $9.0 \ S \ cm^2 \ mol^{-1}$?

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