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A conductivitycell findwith 0.1 M KCIgivesat 25^(@) C aresistance of 85.5 other . Theconductivityof 0.1 M KCIat 25^(@) C is0.01286 "ohm"^(-1) cm^(-1) . Thesamecell filledwith 0.005 M HCIgivesa resistanceof 529ohms.What isthe molarconductivityof HCIsolutionat 25^(@)C?

Answer»


Solution :Given :Resistance Of KCIsolution= `R_(KCI) = 85.5 Omega`
Conductivityof KCIsolution`= k_(KCI) = 0.01286 ohm^(-1)CM^(-4)`
Concentration = C=0.005 M HCI
Resistanceof HCIsolution = `R _(soln) =529OHMS `
Molarconductivity 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)`
Conductivityof HCIsolution
`k_("soln")= ( b )/(R_("soln"))= (1.1)/(529)= 2.08 xx 10^(-3)ohm^(-1) cm^(-1)`
Molarconductivity`^^_(m_(HCI)) = .(k_("soln")xx 1000)/(C )`
`=(2.08 xx 10^(-3) xx 1000)/(0.005)`
`= 416 " ohm"^(-1) cm^(2)mol^(-1)`


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