The specific conductance of a saturated solution of silver bromide is `KScm^(-1)`. The limiting ionic conductivity of `Ag^(+)` and `Br^(-)` ions are x and y respectively. The solubility of silver bromide in g/L is (molar mass of `AgBr=188`)

Options

A. `(Kxx1000)/(x-y)`
B. `(K)/(x+y)xx188`
C. `(Kxx1000xx188)/(x+y)`
D. `(x+y)/(k)xx(1000)/(188)`

Answer

C

Explanation

`AgBr=x+y`
`AgBr=(Kxx1000)/(M)`
M-molarity of `AgBr` solution
8 (solubility in g/L)`=(Kxx1000)/((x+y))xx188`