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The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation `lambda_m^( C)=lambda_m^(oo)-bsqrtC` where `lambda_(m)^( C)`=molar specific conductance `lambda_(m)^(oo)`=molar specific conductance at infinite dilution C=molar concentration `{:("Molar Concentration of NaCl","Molar Conductance" "In" "ohm"^(-1)cm^(2)"mole"^(-1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):}` When a certain conductivity cell (C) was filled with `25xx10^(-4)`(M) NaCl solution.The resistance of the cell was found to be 1000 ohm.At Infinite dilution, conductance of `Cl^(-)` and `SO_4^(-2)` are `80 ohm^(-1) cm^(2) "mole"^(-1) and 160 ohm^(-1) cm^2 "mole"^(-1)` respectively. If the cell ( C) is filled with `5xx10^(-3)(N)Na_(2)SO_(4)` the obserbed resistance was 400 ohm.What is the molar conductance of `Na_(2)SO_(4)`.A. `19.25 "ohm"^(-1)cm^2"mole"^(-1)`B. `96.25 "ohm"^(-1)cm^2"mole"^(-1)`C. `385 "ohm"^(-1)cm^2"mole"^(-1)`D. `192.5 "ohm"^(-1)cm^2"mole"^(-1)` |
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Answer» Correct Answer - D `sigma=(Mlambda)/(1000 G)=(MlambdaR)/1000implies lambda=(1000sigma)/(MR)` `lambda=((1000xx0.1925)/(5xx10^(-)3))/2xx400=192.5` |
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