1.

A conductivity cell find with 0.1 M KCI gives at `25^(@) C` a resistance of 85.5 other . The conductivity of 0.1 M KCI at `25^(@) C` is 0.01286 `"ohm"^(-1) cm^(-1) `. The same cell filled with 0.005 M HCI gives a resistance of 529 ohms. What is the molar conductivity of HCI solution at `25^(@) C`?

Answer» Correct Answer - Molar conductivity of HCI solution . `=^^_(m_((HCI)))`
`=416 ohm^(-1) cm^(2) mol^(-1)`
Given : Resistance Of KCI solution = `R_(KCI) = 85.5 Omega`
Conductivity of KCI solution `= k_(KCI) = 0.01286 ohm^(-1) cm^(-4)`
Concentration = C= 0.005 M HCI
Resistance of HCI solution = `R _(soln) = 529 ohms `
Molar conductivity of HCI `= ^^_(m(HCI)) = ? `
Conductivity `= (" Cell constant ")/("Resistance ")`
`k_(KCI) = ( b) /(R_(KCI))`
`:. b =(k_(KCI)) xx R_(KCI) = 0.01286 xx 85.5 = 1.1 cm^(-1)`
Conductivity of HCI solution
`k_("soln") = ( b )/(R_("soln")) = (1.1)/(529) = 2.08 xx 10^(-3) ohm^(-1) cm^(-1)`
Molar conductivity `^^_(m_(HCI)) = .(k_("soln")xx 1000)/(C )`
`=(2.08 xx 10^(-3) xx 1000)/(0.005)`
`= 416 " ohm"^(-1) cm^(2) mol^(-1)`


Discussion

No Comment Found

Related InterviewSolutions