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What is the molar conductivity of Agl ...

What is the molar conductivity of Agl at zero concentration if the `^^_(0)` values of NaI, `AgNO_(3)` and `NaNO_(3)` are respectively `126.9 Omega^(-1) cm^(2) mol^(-1), 133.4 Omega^(-1) cm^(2) mol^(-1) ` and 121.5 `Omega^(-1) cm^(2) mol^(-1)` ?

Text Solution

Verified by Experts

The correct Answer is:
`^^_(o_((Agl))) = 138.8 Omega^(-1) cm^(2) mol^(-1)`
Given : `^^_(o_((NAI))) = 126. 9 Omega^(-1) cm^(2) mol^(-1)`
` ^^_(o_((AgNO_(2)))) = 133.4 Omega^(-1) cm^(2) mol^(-1)`
`^^_(o_((NaNO_(3)))) = 121. 5 Omega^(-1) cm^(2) mol^(-1)`
By Kohlrausch's law
`^^_(o_((Agl))) = lambda_(Ag^(+))^(o) + lambda_(i^(-))^(0) " "......(I)`
`^^_(o_((NaI)))= lambda_(Ag^(+))^(0) + lambda_(I^(-))^(0) " "......(i)`
`^^_(o_((AgNO_(3))))= lambda_(Ag^(+))^(0) + lambda_(I^(-))^(0)" "......(ii)`
`^^_(o_((NaNO_(3))))= lambda_(Na^(+))^(0) + lambda_(NO_(3))^(3)`
Adding equation (i) and (ii) and sobtracting equation (iii) we get equation I.
`^^_(0_((Agl)))= ^^_(0_((NaI))) + ^^_(0_((AgNO_(3)))).^^(0_((NaNO_(3)))`
`= 126.9 + 133.4 - 121 .5 `
`=138. 8 Omega^(-1) cm^(2) mol^(-1)`
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