Molar conductivity of a solution is `1.26xx10^(2)` `ohm^(-1)` `cm^(2)` `mol^(-1)` . Its molarity is 0.01. Its specific conductivity will be

Options

A. `1.26xx10^(-25)` `ohm^(-1)` `cm^(-1)
B. `1.26xx10^(-3)` `ohm^(-1)` `cm^(-1)`
C. `1.26xx10^(-4)` `ohm^(-1)` `cm^(-1)`
D. 0.0063 `ohm^(-1)` `cm^(-1)`

Answer

B

Explanation

Applying, `Lambda_m=kxx1000/"Molarity"`
`k=(1.26xx10^(2)xx0.01)/1000=1.26xx10^(-3) ohm^(-1) cm^(-1)`