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The solubility product, $K_{sp}\: of Ag_2\: CrO_4$ is $ 3.2 \times 10^{-11}\: at T(K)$ . Its solubility in mol $ L^{-1}$ is

$\begin {array} {1 1} (1)\;4 \times 10^{-4} & \quad (2)\;2 \times 10^{-4} \\ (3)\;2 \times 10^{-3} & \quad (4)\;4 \times 10^{-6} \end {array}$

 

1 Answer

(2) $2 \times 10^{-4}$
answered Nov 7, 2013 by pady_1
 
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