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^-6`
B. `2.5 xx 10^-9`
C. `2.5 xx 10^-13`
D. `4 xx 10^-12`

Answer

A

Explanation

`lamda_m = (1000 K)/(S) = (1000 xx 3.06 xx 10^-6)/(S) = 1.53`
`S = 2 xx 10^-3 (mol)/(litre)`
`K_(sp)(BaSO_4)` = `S^2 = (2 xx 10^-3)^2 = 4 xx 10^-6`.