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_
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)`
1 Answers
Correct Answer - D
`sigma=(Mlambda)/(1000 G)=(MlambdaR)/1000implies lambda=(1000sigma)/(MR)`
`lambda=((1000xx0.1925)/(5xx10^(-)3))/2xx400=192.5`