If the solubility of a sparingly soluble salt of the type $B{A_2}$ (giving three ions on dissociation of a molecule) is $x$ moles per litre, then its solubility product is given by

The $2.901$ $L$ saturated solution is formed by $0.08$ $g$ $Ca{F_2}$ at $298$ $K$ temp. then calculate ${K_{sp}}$

( Molecular moles of $Ca{F_2}$ is $78$ $g$ $mol^{-1}$ )

In $1 L$ saturated solution of $AgCl \left[ K _{ sp }( AgCl )=1.6 \times 10^{-10}\right]$, $0.1$ mol of CuCl $\left[ K _{ sp }( CuCl )=1.0 \times 10^{-6}\right]$ is added. The resultant concentration of $Ag ^{+}$in the solution is $1.6 \times 10^{-x}$. The value of " $x$ " is

Solubility product expression of salt $MX_4$ which is sparingly soluble with a solubility $(S)$ can be given as

The solubility product constant of $Ag _{2} Cr O _{4}$ and $AgBr$ are $1.1 \times 10^{-12}$ and $5.0 \times 10^{-13}$ respectively. Calculate the ratio of the molarities of the molarities of their saturated solutions.