The conductivity of a saturated solution of `BaSO_4` is ` 3. 06 xx 10^(-6) "ohm"^(-1) cm^(-1)` and its equivalent conductance is `1.53 "ohm"^(-1) cm^2 equiv^(-1)`. The `K_(sp)` for `BaSO_4` will be .

Options

A. ` 4 xx 10^(-12)`
B. `2.5xx10^(13)`
C. `25xx10^9`
D. `10^(-6)`

Answer

D

Explanation

` 1.53 = (1000 xx 3 .06 xx 19^(-6))/("Normality")`
Normality `= 2 xx 10^(-3) N`
Molaroty `= (2 xx 10^(-3))/2 =10^(-3) M`
`K_(sp) = 10^(-6) M`.