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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“. |
| 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∞(NH4OH)=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∞(NH4OH)=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=502.88cm2mol6−1Molar conductivity at infinite dilution for NH_4OHNH4OH is 523.28 cm^2mol^{-1}mol−1 . | |