The molar conductivities of KCl,NaCl and KNO_ | vedclass

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(A) According to Kohlrausch law,the molar conductivity of an electrolyte can be expressed as the sum of the molar conductivities of its constituent io

Vedclass · Accepted Answer

$110$

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(A) According to Kohlrausch law,the molar conductivity of an electrolyte can be expressed as the sum of the molar conductivities of its constituent ions.$\Lambda_{m}^{\circ}(NaNO_3) = \lambda^{\circ}(Na^+) + \lambda^{\circ}(NO_3^-)$We can express this in terms of the given electrolytes:$\Lambda_{m}^{\circ}(NaNO_3) = \Lambda_{m}^{\circ}(NaCl) + \Lambda_{m}^{\circ}(KNO_3) - \Lambda_{m}^{\circ}(KCl)$Substituting the given values:$\Lambda_{m}^{\circ}(NaNO_3) = 120 + 90 - 100$$\Lambda_{m}^{\circ}(NaNO_3) = 210 - 100 = 110 \ S \ cm^2 \ mol^{-1}$.

The molar conductivities of $KCl$,$NaCl$ and $KNO_3$ are $100$,$120$ and $90 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of $NaNO_3$ would be .......... $S \ cm^2 \ mol^{-1}$.

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