The solubility product of $PbI_2$ is $1.08 \times 10^{-7}$. Calculate its solubility in $mol \ dm^{-3}$ at $298 \ K$?

If $K_{sp}$ of $CaF_2$ at $25\,^{\circ}C$ is $1.7 \times 10^{-10}$,the combination amongst the following which gives a precipitate of $CaF_2$ is

Predict whether a precipitate of $PbI_2$ will be formed or not on mixing $20 \ mL$ of $3 \times 10^{-3} \ M$ $Pb(NO_3)_2$ solution with $80 \ mL$ of $2 \times 10^{-3} \ M$ $NaI$ solution. $K_{sp}$ for lead iodide $(PbI_2)$ is $6.0 \times 10^{-9}$.

The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility in $mol/L$ is:

Assign $A, B, C, D$ from the given type of reaction: $PbCl_2(s) + \text{Hot water} \longrightarrow Pb^{2+}(aq) + 2Cl^{-}(aq)$

The $K_{sp}$ of $CaF_2$ is $1.7 \times 10^{-10}$. What is the volume in $mL$ of a saturated solution containing $10 \ mg$ of $CaF_2$? (Molecular mass of $Ca = 40, F = 19$).