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The limiting molar conductivities of NaI, NaNO3 and AgNO3 are 12.7, 12.0 and 13.3 mS m2 mol−-−1, respectively (all at 25∘^\circ∘C). The limiting molar conductivity of AgI at this temperature is ____________ mS m2 mol−-−1.
Given
(1) λm∞(NaI)=12.7 mS m2 mol−1\lambda_m^{\infty}(\mathrm{NaI})=12.7 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}λm∞(NaI)=12.7 mS m2 mol−1
(2) λm∞(NaNO3)=12.0 mS m2 mol−1\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NaNO}_3\right)=12.0 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}λm∞(NaNO3)=12.0 mS m2 mol−1
(3) λm∞(AgNO3)=13.3 mS m2 mol−1\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{AgNO}_3\right)=13.3 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}λm∞(AgNO3)=13.3 mS m2 mol−1
λm∞(Ag I)=(1)+(3)−(2)\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=(1)+(3)-(2)λm∞(Ag I)=(1)+(3)−(2)
=12.7+13.3−12.0 =12.7+13.3-12.0 =12.7+13.3−12.0
=26.0−12.0 =26.0-12.0 =26.0−12.0
λm∞(Ag I)=14.0\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=14.0λm∞(Ag I)=14.0
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