Solubility of $Ca_{3}(PO_{4})_{2}$ is $S \ mol \ dm^{-3}$. Find the solubility product $(K_{sp})$.

The solubility product of a binary weak electrolyte is $4 \times 10^{-10}$ at $298 \ K$. Its solubility in $\text{mol} \ dm^{-3}$ at the same temperature is

The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

The solubility products of some sparingly soluble electrolytes are given below. The increasing order of their molar solubility is:
$(a)$ Formula: $MX$,$K_{sp} = 4.0 \times 10^{-20}$
$(b)$ Formula: $P_2Q$,$K_{sp} = 3.2 \times 10^{-11}$
$(c)$ Formula: $LY_3$,$K_{sp} = 2.7 \times 10^{-31}$

The solubility of calcium carbonate $(CaCO_3)$ at $298 \ K$ is $6.4 \times 10^{-5} \ mol \ dm^{-3}$. Calculate the value of solubility product $(K_{sp})$ at the same temperature.

At $T(K)$ in a saturated solution of $MgCO_3$ and $Ag_2CO_3$, if the concentration of $Mg^{2+}$ ion is $3.2 \times 10^{-5} \ M$, then the concentration of $Ag^{+}$ ion in the solution will be [Given: $K_{sp}(MgCO_3) = 1.6 \times 10^{-6}$ and $K_{sp}(Ag_2CO_3) = 8.0 \times 10^{-12}$ at $T(K)$]