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At infinite dilution, molar conductivities of Ba(OH)_(2), BaCl_(2) and NH_(4)Cl solution are 523.28, 280.0 and "129.8 ohm"^(-1)."cm"^(2)."mol"^(-1) respectively. Calculate the molar conductivity of NH_(4)Cl solution at infinite dilution. |
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Answer» Solution :`Lambda_(m)^(@)[BA(OH)_(2)]=lambda_(m)^(@)(B^(+2))+2lambda_(m)^(@)(OH^(-))` `=523.28" OHM"^(-1)."cm"^(2)."mol"^(-1)"...[1]"` `Lambda_(m)^(@)(BaCl_(2))=lambda_(m)^(@)(Ba^(+2))+2lambda_(m)^(@)(Cl^(-))` `="280.0 ohm"^(-1)."cm"^(2)."mol"^(-1)"...[2]"` and `Lambda_(m)^(@)(NH_(4)Cl)=lambda_(m)^(@)=lambda_(m)^(@)(NH_(4)^(+))+lambda_(m)^(@)(Cl^(-))` `="129.8 ohm"^(-1)."cm"^(2)."mol"^(-1)...[3]"` Subtracting equation [2] from [1] gives `2lambda_(m)^(@)(OH^(-))-2lambda_(m)^(@)(Cl^(-))="243.28 ohm"^(-1)."cm"^(2)."mol"^(-1)"...[4]"` Multiplying equation [3] by 2 and then adding the remit to equation [4], we get `2lambda_(m)^(@)(NH_(4)^(+))+2lambda_(m)^(@)(OH^(-))` `=2xx129.8+243.28=502.88" ohm"^(-1)."cm"^(2)."mol"^(-1)` `therefore""Lambda_(m)^(@)(NH_(4)OH)=(502.88)/(2)="251.44 ohm"^(-1).cm"^(2)."mol"^(-1)` |
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