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At infinite dilution, molar conductiviti...

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.

Text Solution

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`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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