The equivalent conductivity of `H_(2)SO_(4)` at infinite dilution is `384 Omega^(-1) cm^(2) eq^(-1)`. If `49g H_(2)SO_(4)` per litre is present in solution and specific resistance is `18.4 Omega` then calculate the degree of dissociation.

Explanation

Equivalent of `H_(2)SO_(4) = (49)/(49) = 1N`
Specific conductance `= (1)/("specific resistance") = (1)/(18.4)`
`rArr lambda_(eq) = (100 xx k)/(N) = (1000 xx 1)/(18.4) = 55`
Degree of dissociation `(alpha) = (lambda_(eq.)^(c))/(lambda_(eq.)^(prop)) = (55)/(384)`
`= 0.14 rArr alpha % = 14%`