At `25^(@)C`, the resistance of `0.01 N NaCl` solution is `200 ohm` . If cell constant of the conductivity cell is unity, then the equivalent conductance of the solution is:

Options

A. `400 S cm^(2) eq^(-1)`
B. `900 S cm^(2) eq^(-1)`
C. `500 S cm^(2) eq^(-1)`
D. `0.05 S cm^(2) eq^(-1)`

Answer

C

Explanation

`:' Lambda_(eq.) = kappa xx (1000)/(N) = (1000)/(N) xx (1)/(R) xx (1)/(a)`
`= (1000)/(0.01) xx (1)/(200) xx 1 = 500 S cm^(2) eq^(-1)`