The osmotic pressure of a saturated aqueous solution of an entirely dissociated salt type $ M_2X_3$ is worth $3.0\;atm $ at $300\;K$.
Then calculate the solubility product $K_s$!
$i=5.0$
$3.0=5.0 [M_2X_3]0.082\cdot 300$
$[M_2X_3]=0.024\frac{mol}{L}$
By ionisation, one mole produces 2 moles $M^{3+}$ and 3 moles $X^{2-}$:
$K_s$ $=$ $[M^{3+}]^2[X^{2-}]^3$ $=$ $0.048^2\cdot 0.072^3$ $=$ $8.6\cdot 10^{-7}\cdot M^5$
