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Q.
Solubility of calcium phosphate (molecular mass, $M$ ) in water is $W$ g per $100\, mL$ at $25^{\circ} C$. Its solubility product at $25^{\circ} C$ will be approximately.
Equilibrium
Solution:
$\because S=\frac{10 W}{M} mol / L$
For $Ca _{3}\left( PO _{4}\right)_{2} \rightleftharpoons \underset{3S}{3 Ca ^{2+}}+\underset{2S}{2 PO _{4}^{3-}}$
$\Rightarrow K_{ sp }=(3 S)^{3} \times(2 S)^{2}$
$\therefore K$ of $Ca _{2}\left( PO _{4}\right)_{2}=108 S^{5}$
$\Rightarrow 108\left(\frac{10 W}{M}\right)^{5}$
$\Rightarrow 10^{7}\left(\frac{W}{M}\right)^{5}($ approximately $)$