The solubility product of $Mg(OH)_2$ is $1.2 \times 10^{-11}$. The solubility of this compound in gram per $100 \ cm^3$ of solution is

The required amount of $KBr$ (molar mass $= 119 \ g/mol$) in grams to start the precipitation of $AgBr$ in $500 \ mL$ solution of $0.05 \ M \ AgNO_3$ will be :- ($K_{SP}$ of $AgBr = 5 \times 10^{-13}$)

Calculate the solubility of sparingly soluble salt $BA$ in $mol \ dm^{-3}$ at $300 \ K$ if its solubility product is $4.9 \times 10^{-9}$ at same temperature.

The solubility of $AgCl$ at $20 \, ^oC$ is $1.435 \times 10^{-3} \, g/L$. The solubility product $(K_{sp})$ of $AgCl$ is:

Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.

The solubility products of three sparingly soluble salts are given below. What is the correct decreasing order of their molar solubility?

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$