`0.04 N` solution of a weak acid has specific conductance `4.23 xx 10^(-4) "mho cm"^(-1)`. If the degree of dissociation of acid at this dilution is `0.0612`, then equivalent conductivity at infinite dilution is :A. `150.8 "mho cm"^(2) eq^(-1)`B. `172.8 "mho cm"^(2) eq^(-1)`C. `180.6 "mho cm"^(2) eq^(-1)`D. `160.9 "mho cm"^(2) eq^(-1)`

`0.04 N` solution of a weak acid has specific conductance `4.23 xx 10^

1.	`0.04 N` solution of a weak acid has specific conductance `4.23 xx 10^(-4) "mho cm"^(-1)`. If the degree of dissociation of acid at this dilution is `0.0612`, then equivalent conductivity at infinite dilution is :A. `150.8 "mho cm"^(2) eq^(-1)`B. `172.8 "mho cm"^(2) eq^(-1)`C. `180.6 "mho cm"^(2) eq^(-1)`D. `160.9 "mho cm"^(2) eq^(-1)`
Answer» Correct Answer - B `Lambda_(v) = kappa xx (1000)/(N)` `= 4.23 xx 10^(-4) xx (1000)/(0.04) = 10.575` Now `alpha = (Lambda_(v))/(Lambda_(oo))` `:. (Lambda_(oo)) = (10.575)/(0.0612) = 172.79`

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