The molar conductivity of a `0.5 mol//dm^(3)` solution of `AgNO_(3)` with electrolytic conductivity of `5.76 xx 10^(-3) S cm^(-1)` at `298 K` is

Options

A. `2.88" S "cm^(2)//mol`
B. `11.52" S "cm^(2)//mol`
C. `0.086" S "cm^(2)//mol`
D. `28.8" S "cm^(2)//mol`

Answer

B

Explanation

`wedge_(m)=(kappaxx1000)/("Molarity")`
`=((5.76xx10^(-3)" S "cm^(-1))(1000cm^(3)L^(-1)))/(0.5" mol "L^(-1))`
`=(5.76)/(0.5)=11.52" S "cm^(2)mol^(-1)`.