if `^^_(c)` of `NH_(4)OH` is `11.5Omega^(-1) cm^(2) mol^(-1)`, its degree of dissociation would be
(Given. `lambda_(NH_(4)^(+))^(infty)=73.4Omega^(-1) cm^(2) mol^(-1)` and `lambda_(OH^(-))^(infty)=197.6Omega^(-1) cm^(2) mol^(-1))`

The molar conductivity of 0.025 M methanoic acid (HCOOH) is 46.15" S "cm^(2)mol^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1) and lambda_((HCOO^(-)))^(@)=54.6" S " cm^(2)mol^(-1) .

The molar conductivity of 0.25 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Calculate the degree of dissociation constant. Given : lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) and lambda_(HCOO^(-))^(@)=54.6Scm^(2)mol^(-1)

The conductivity of 0.001 " mol "L^(-1) solution of CH_(3)COOH is 4.95xx10^(-5)" S "cm^(-1) . Calculate its molar conductance and degree of dissociation (alpha). "Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)

The value of ^^_(m)^(infty) for KCl and KNO_(3) are 149.86 and 154.96 Omega^(-1) cm^(2) mol^(-1) Also lambda_(Cl^(-))^(infty) is 71.44 " ohm"^(-1) cm^(2)mol^(-1) The value of lambda_(NO_(3)^(-))^(infty) is

The conductivity of 0.20 mol L^(-1) solution of KCI is 2.48xx10^(-2)S cm^(-1) . Calculate its molar conductivity and degree of dissociation (alpha) . Given lambda_((K^(+)))^(@)=73.5 S cm^(-2)mol^(-1)and lambda_((CI^(-)))^(@)=76.5 mol^(-1) lambda_((K^(+)))^(@)=73.5 S cm^(-2)mol^(-1)and lambda_((CI^(-)))^(@)=76.5 mol^(-1)

The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Its degree of dissociation (alpha) and dissociation constant. Given lambda^(@)(H^(+))=349.6 S cm^(-1) and lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Its degree of dissociation (alpha) and dissociation constant. Given lambda^(@)(H^(+))=349.6 S cm^(-1) and lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .