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 direction C=Molar concentration `{:("Molar Concentration of NaCl","Molar Conductance inohm"^(-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 800 `"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_2SO_4` the observed resistance was 400 ohm. What is the molar conductance of `Na_2SO_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)`

The molar conductance of NaCl varies with the concentration as shown i

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 direction C=Molar concentration `{:("Molar Concentration of NaCl","Molar Conductance inohm"^(-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 800 `"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_2SO_4` the observed resistance was 400 ohm. What is the molar conductance of `Na_2SO_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)`
Answer» Correct Answer - D For `Na_2SO_4` solution `K=(l/a)xx1/R=0.1925/400=4.81xx10^(-4) ohm^(-1)cm^(-1)` `lambda_m=(Kxx1000)/M=(4.81xx10^(-4)xx1000)/(5/2xx10^(-3))` `lambda_m(Na_2SO_4)=192.4 ohm^(-1) cm^2 "mole"^(-1)`

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