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From the following molar conductivities ...

From the following molar conductivities at infinite dilution.
`Lambda_(m)^(@) ` for `Ba(OH)_(2) = 457.6 Omega^(-1) cm^(2) "mol"^(-1)`
`Lambda_(m)^(@) ` for `BaCl_(2) = 240.6 Omega^(-1) cm^(2) "mol"^(-1)`
`Lambda_(m)^(@) ` for `NH_(4) Cl = 129.8 Omega^(-1) cm^(2) "mol"^(-1)`
Calculate `Lambda_(m)^(@)` for `NH_(4) OH`.

Text Solution

Verified by Experts

`Lambda_(m)^(@) [ Ba( OH)_(2)] = lambda_(Ba^(2))+ 2 lambda_(OH^(-))^(@)`
`= 457.6 S cm^(2) "mol"^(-1)`
`Lambda_(m)^(@) ( BaCl_(2)) = lambda_(Br^(2+))^(@) + 2 lambda_(Cl^(-))^(@) = 240.6 S cm^(2) "mol"^(-1)`
`Lambda_(m)^(@) ( NH_(4) Cl) = lambda_(NH_(4))^(@) + lambda_(Cl^(-))^(@) = 129.8 S cm^(2) "mol"^(-1)`
`Lambda_(m)^(@) ( NH_(4) OH) = Lambda_(m)^(@) ( NH_(4) Cl) + 1//2Lambda_(m)^(@) [ Ba(OH)_(2) ] - 1//2 Lambda_(m)^(@) [ BaCl_(2)]`
`129.8 + 1//2( 457.6) - 1//2 ( 240.6)`
= 238.3 S `cm^(2) "mol"^(-1)`
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