InterviewSolution
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InterviewSolution
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Resistance of 0.2 M solution of an electrolyte is 50 `Omega`. The specific conductance of the solution is 1.4 S `m^(-1)`. The resistance of 0.5 M solution of the same electrolyte is `280Omega`. The molar conductivity of 0.5 M solution of the electrolyte in S `m^(2)mol^(-1)` isA. `5xx10^(2)`B. `5xx10^(-4)`C. `5xx10^(-3)`D. `5xx10^(3)` |
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Answer» Correct Answer - B Case I: `C=0.2M,R=50Omega,kappa=1.4" S "cm^(-1)` `kappa`=Conductance`xx`Cell const.`=(1)/(R)xx`cell const. ltBrgt `therefore`Cell constant`=kappaxxR=(1.4" S "m^(-1))(50Omega)=70m^(-1)` Case II: `C=0.5" M ",R=280Omega," cell cont."=70m^(-1)` `kappa=(1)/(R)xx"cell constant"=(1)/(280Omega)xx70m^(-1)` Molar conductivity (in S `m^(2)mol^(-1)`) `=(kappa(S" "m^(-1)))/("Molarity"(mol " "L^(-1))xx1000" L "m^(-3))` `=(0.25" S "m^(-1))/(0.5 " mol "L^(-1)xx1000" L "m^(-3))` `=5xx10^(-4)" S "m^(2)mol^(-1)` |
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