Electrolytic conductivity of 0.30 M solution of KCl at 298 K is `3.72xx10^(2) " S "cm^(-1)`. Calculate its molar conductivity.

Explanation

Conductivity (k)`=3.72xx10^(-2) S cm^(-1)=3.72xx10^(-2) "ohm"^(-1) cm^(-1)`
Molar concentration (C )`=0.3 M =0.3 "mol "L^(-1)=(0.3 " mol")/(1000 cm^(3))=3.0xx10^(-4) "mol" cm^(-3)`
`"Molar conductance" (Lambda_(m))=(k)/(C )=((3.72xx10^(-2) "ohm"^(-1) cm^(-1)))/((3.0xx10^(-4) "mol" cm^(3)))=124" ohm"^(-1) mol^(-1) cm^(2)`
`=124 S "mol"^(-1) cm^(2)`