1.

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).

Answer»

`19.25 "OHM"^(-1)cm^2"mole"^(-1)`
`96.25 "ohm"^(-1)cm^2"mole"^(-1)`
`385 "ohm"^(-1)cm^2"mole"^(-1)`
`192.5 "ohm"^(-1)cm^2"mole"^(-1)`

Solution :`SIGMA=(Mlambda)/(1000 G)=(MlambdaR)/1000implies lambda=(1000sigma)/(MR)`
`lambda=((1000xx0.1925)/(5XX10^(-)3))/2xx400=192.5`


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