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  • The limiting molar conductivities for <img alt="\mathrm{NaCl}, \mathrm{KBr} \mathr

The limiting molar conductivities \Lambda^{\circ} for \mathrm{NaCl}, \mathrm{KBr} \mathrm{~and} \mathrm{~KCl} are 142,130 and 120 \mathrm{Scm}^2 \mathrm{~mol}^{-1} respedicly. Calculate the x for NaBr.

Option: 1

268


Option: 2

177


Option: 3

152


Option: 4

124


Answers (1)

best_answer

\Lambda^{\circ}{\mathrm{NaCl}}=\lambda^{\circ} \mathrm{Na}^{+}+\lambda \mathrm{Cl} \\     .....(1)

\Lambda^{\circ} \mathrm{KBr}=\lambda \mathrm{K}^{+}+\lambda \mathrm{Br}^{-} \         .....(2)

\Lambda^{\circ} \mathrm{KCl}=\lambda \mathrm{K}^{+}+\lambda \mathrm{Cl}^{-}          .....(3)

By (1), (2) & (3)

\Lambda_{\mathrm{NaBr}} =\lambda \mathrm{Na}^{+}+\lambda{\mathrm{Br}^{-}} \\

=142+130-120 \\

=152 \mathrm{~scm}^2 \mathrm{~mol}^{-1}

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