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Question
The conductivity of 0.20 mol L−1 solution of KCl is 2.48 × 10−2 S cm−1. Calculate its molar conductivity and degree of dissociation (α). Given λ0 (K+) = 73.5 S cm2 mol−1 and λ0 (C1−) = 76.5 S cm2 mol−1.
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Solution
Conductivity of KCl solution = 2.48×10–2 S cm–1
Concentration of KCl solution = 0.20 mol L–1
= 0.20 × 1000 mol cm–3
= 200 mol cm–3
Molar conductivity
`Lambda_m=k/c`
`=(2.48xx10^(-2)" S "cm^(-1))/(200" mol "cm^(-3))`
`=124xx10^(-6)" S "cm^2" mol"^(-1)`
Given:
`lambda_(K^+)^@=73.5" S "cm^2" mol"^(-1)`
`lambda_(Cl^-)^@=76.5" S "cm^2" mol"^(-1)`
Degree of dissociation
`alpha=Lambda_m/Lambda_m^@" "" "" "" ".......("i")`
`Lambda_(m(KCl))^@=lambda_(K+)^@+lambda_(Cl^-)^@`
`=73.5+76.5`
`=150" S "cm^2" mol"^(-1)`
`"Substituting the values of "Lambda_m" and "Lambda_(m(KCl))^@" in (i), we get"`
`alpha=(124xx10^(-6)" S "cm^2" mol"^(-1))/(150" S "cm^2" mol"^(-1))=0.82xx10^(-6)`
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