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The molar conductivity at zero concen...

The molar conductivity at zero concentration of `NH_(4) CI` NaOH and NaCI are respectively 149.7`Omega^(-1) cm^(2) mol^(-1) 248.1 Omega^(-1) cm^(2) mol^(-1) ` and `126.5 Omega^(-1) cm^(2) mol^(-1).` What is the molar conductivity of `NH_(4)OH` at zero concentration ?

Text Solution

Verified by Experts

The correct Answer is:
`^^_(o_((NH_(4)OH))=271.3 Omega^(-1) cm^(2) mol^(-1)`
`^^_(o_((NH_(4)C1)) = 149.1 Omega^(-1) cm^(2) mol^(-1)`
`^^_(o_((NaOH)))=248.1 Omega^(-1) cm^(2) mol^(-1)`
`^^_(o_((NaCI)))=126.5 Omega^(-1) cm^(2) mol^(-1)`
By Kohlrausch's law
`^^_(o_((NH_(4)OH)))= lambda_(NH_(4))^(0) + lambda_(OH^(-))^(0)" "......(I)`
`^^_(o_((NH_(4)C1)))=lambda_(NH_(4))^(0) + lambda_(C1)^(0)" "......(i)`
`^^_(o_((NaOH)))=lambda_(Na^(+))^(0) + lambda_(OH^(-))^(0) " "......(ii)`
`^^_(0_((NaC1)))=lambda_(Na^(+))^(0) + lambda _(C1^(-))^(0)" "......(iii)`
Adding equations (i) and (ii) and subtracting equation (iii) we get equation I.
`:. ^^_(o_((NH_(4)OH))) = ^^_(o_((NH_(4)C1))) + ^^_(0_((NaOH))) ^(-) ^^_(0_((NaC1)))`
`=149.7 + 248 .1- 126. 5`
`=271.3 Omega^(-1) cm^(2) mol^(-1)`
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