At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl is :

Options

A. `10^(5) cm^(3)`
B. `10^(3) cm^(3)`
C. `10 cm^(3)`
D. `10^(5) cm^(2)`

Answer

C

Explanation

`lambda = (K xx 1000)/C`
`lambda/K = 10^(3)/0.01 = 10^(5) cm^(3) eq^(-1) , therefore lambda = (Omega^(-1) cm^(2) eq^(-1))/(Omega^(-1) cm^(-1))= cm^(3) eq^(-1)`