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. What is the cell constant of the conductivity cell ( C) ?A. `0.385 cm^(-1)`B. `3.85 cm^(-1)`C. `38.5 cm^(-1)`D. `0.1925 cm^(-1)`

Answer» Correct Answer - D
For `25xx10^(-4)` (M) NaCl solution
`lambda_m=lambda_m^oo-bsqrtC`
`lambda_m=127-10^3(25xx10^(-4))^(1//2)`
`lambda_m=127-10^(3)xx5xx10^(-2)`
`lambda_m=77`
But `lambda_m=(Kxx1000)/M " " K=(l/a)xx1/R`
`lambda_m=(l/a)xx1/Rxx1000/M`
`lambda_m`=[Cell constant]`xx1000/(RxxM)`
`implies `77=[Cell constant]`xx1000/(1000xx25xx10^(-4))`
Cell constant `=77xx25xx10^(-4)=0.1925 cm^(-1)`


Discussion

No Comment Found

Related InterviewSolutions