Question

A
Q.2 Calculate ^° for NHA OH from the following data
°(Ba(OH2)]=457.6 S cm’mor'. ^ °[Bacl2)=240.6 S cm’mor' and
^ °(NH4cl]=129.8 Scm? mor“.
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Answer 1

Answer:

Molar conductivity at infinite dilution for NH_4OHNH4OH is 523.28 cm^2 mol^{-1}cm2mol−1 .

Explanation:

\Lambda _m^{\infty}(NH_4OH)=\lambda_m^{\infty}

(NH_4^{+})+\lambda_m^{\infty}(OH^{-})=?Λm∞(NH4

OH)=λm∞(NH4+)+λm∞(OH−)=?

\Lambda _m^{\infty}(Ba(OH)_2)=\lambda_m^{\infty}

(Ba^{2+})+\lambda_m^{\infty}(OH^{-})= 523.28

cm^2 mol^{-1}Λm∞(Ba(OH)2)=λm∞(Ba2+)+λm∞

(OH−)=523.28cm2mol−1 ..(1)

\Lambda _m^{\infty}(Ba(Cl)_2)=\lambda_m^{\infty}

(Ba^{2+})+\lambda_m^{\infty}(Cl^{-})= 280

cm^2 mol^{-1}Λm∞(Ba(Cl)2)=λm∞(Ba2+)+λm∞

(Cl−)=280cm2mol−1 ..(2)

\Lambda _m^{\infty}(NH_4Cl)=\lambda_m^{\infty}

(NH_4^{+})+\lambda_m^{\infty}(Cl^{-})= 129.8

cm^2 mol^{-1}Λm∞(NH4Cl)=λm∞(NH4+)+λm∞(

(Cl−)=129.8cm2mol−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^{-})Λm∞(NH4

OH)=λm∞(Ba2+)+λm∞(OH−)−λm∞(Ba2+)−λm∞(Cl−)

+2λm∞(NH4+)+2λm∞(Cl−)=λm∞(NH4+)+λm∞(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}λm∞(NH4+)+λm∞(OH−)=523.28−280+2×129.8cm2mol−1=502.88cm2mol6−1

Molar conductivity at infinite dilution for NH_4OHNH4OH is 523.28 cm^2mol^{-1}mol−1 .