Question

Calculate molar conductivity at infinite dilution for NH4OH ,given that molar conductivities for Ba(OH)2 ,BaCl2 and NH4CL are 523.28, 280 and 129.8s cm^2mol^-1 respectively,

Answer 1

Answer: Molar conductivity at infinite dilution for $NH_4OH$ is 523.28 $cm^2 mol^{-1}$.

Explanation:

$\Lambda _m^{\infty}(NH_4OH)=\lambda_m^{\infty}(NH_4^{+})+\lambda_m^{\infty}(OH^{-})=?$

$\Lambda _m^{\infty}(Ba(OH)_2)=\lambda_m^{\infty}(Ba^{2+})+\lambda_m^{\infty}(OH^{-})= 523.28 cm^2 mol^{-1}$..(1)

$\Lambda _m^{\infty}(Ba(Cl)_2)=\lambda_m^{\infty}(Ba^{2+})+\lambda_m^{\infty}(Cl^{-})= 280 cm^2 mol^{-1}$..(2)

$\Lambda _m^{\infty}(NH_4Cl)=\lambda_m^{\infty}(NH_4^{+})+\lambda_m^{\infty}(Cl^{-})= 129.8 cm^2 mol^{-1}$..(3)

$\Lambda _m^{\infty}(NH_4OH)=\lambda_m^{\infty}(Ba^{2+})+\lambda_m^{\infty}(OH^{-})-\lambda_m^{\infty}(Ba^{2+})-\lambda_m^{\infty}(Cl^{-})+2\lambda_m^{\infty}(NH_4^{+})+2\lambda_m^{\infty}(Cl^{-})=\lambda_m^{\infty}(NH_4^{+})+\lambda_m^{\infty}(OH^{-})$

$\lambda_m^{\infty}(NH_4^{+})+\lambda_m^{\infty}(OH^{-})=523.28-280+2\times 129.8 cm^2mol^{-1}=502.88 cm^2 mol6{-1}$

Molar conductivity at infinite dilution for $NH_4OH$ is 523.28 cm^2$mol^{-1}$.

Answer 2

Answer:

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Explanation:

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