The conductivity of a saturated solution of a sparingly soluble salt, MX, is found to be `4.0 × 10^(-5) Ohm^(-1) cm^(-1)` If `lambda_(m)^(infty) (1/2M^(2+))=50.0 Omega^(-1) cm^(2) mol^(-1)` and `lambda_(m)^(2+)(X^(-)=50 Omega^(-1) cm^(2) mol^(-1))` the solubility product of the salt is about

Options

A. `2xx10^(-10)`
B. `2xx10^(-12)`
C. `32xx10^(-12)`
D. `16xx10^(-8)`

Answer

B

Explanation

`wedge_(m)^(@)(MX_(2))=wedge_(m)^(@)(M^(2+))+2wedge_(m)^(@)(X^(-))=2wedge_(m)^(infty)(1/2M^(2+)+2wedge_(m)^(infty)(X^(-))`
`=(2xx50+2xx50)s cm^(2) mol^(-1) =200 s cm^(2) mol^(-1)`
`k_(sp)=[M^(2+)]{[X^(-)]^(2)=(2xx10^(-4) mol "dm"^(-3)) (4xx10^(-4) mol "dm"^(-3^(2)) =32 xx10^(-12)`