The solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$. The value of $K_{sp}$ of the electrolyte is:

The solubility product of a sparingly soluble salt $A_2B$ is $3.2 \times 10^{-11}$. Its solubility in $mol \cdot L^{-1}$ is

If the solubility products of $AgCl$ and $AgBr$ are $1.0 \times 10^{-10}$ and $3.5 \times 10^{-13}$ respectively,then the relation between the solubilities (denoted by the symbol $S$) of these salts can correctly be represented as:

The solubility of $AgBr_{(s)}$,having solubility product $5 \times 10^{-10}$ in $0.2 \ M$ $NaBr$ solution,equals

$A$ precipitate of $CaF_{2}$ $(K_{sp} = 1.7 \times 10^{-10})$ will be obtained when equal volumes of the following are mixed:

The molar solubility $(s)$ of zirconium phosphate with molecular formula $(Zr^{4+})_3(PO_4^{3-})_4$ is given by the relation: