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** Q 1210423319. ** Calculate Its solubility product at `298 K, `

At `298 K`, the conductivity of a saturated solution of `AgCl` in water is `2.6 × 10^(-6) S cm^(-1)`.

( given :` λ^(∞) (Ag^+ ) = 63.0 S cm^2mol^(-1) , λ^(∞) (Cl^(∞)) = 67.0 S cm^(2)Mol^(-1) `)

#### HINT

Use formula: Solubility `S =(1000k)/(λ_(AgCl)^(0))`